Linear Algebra Applications Nonlinear Analysis and 130 its (2016) 396–407 466 (2015) 102–116
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Linear Algebra and its Applications Nonlinear Analysis www.elsevier.com/locate/laa www.elsevier.com/locate/na
Inverse eigenvalue problem of Jacobi Stable estimates for source solution of critical fractalmatrix with mixed data Burgers equation ∗ Ying Wei 1 Tomasz Jakubowski , Grzegorz Serafin Wroclaw University of Department Technology, of ul.Mathematics, Wybrze˙ze Wyspia´ nskiego 27, Wroclaw, Poland and Astronautics, Nanjing University of Aeronautics Nanjing 210016, PR China
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Article history: In this paper, we provide two-sided estimates for the source solution of d-dimensional α/2 q ) = 0,problem Article history: Received 25 February 2015 this paper, the inverse eigenvalue of reconstructing critical fractal BurgersIn equation ut −∆ +b·∇(u|u| q = (α−1)/d, α ∈ (1, 2), Received 16 January b2014 Accepted 14 October 2015 Jacobi of matrix from itsα-stable eigenvalues, its leading principal ∈ Rd , by the densityafunction the isotropic process. Accepted 20 September 2014 Communicated by Enzo Mitidieri submatrix and part© of theElsevier eigenvalues of rights its submatrix 2015 Ltd. All reserved. Available online 22 October 2014 is considered. The necessary and sufficient conditions for Submitted by Y. Wei Keywords: the existence and uniqueness of the solution are derived. Fractional Laplacian Furthermore, a numerical algorithm and some numerical MSC: Critical Burgers equation examples are given. 15A18 Source solution © 2014 Published by Elsevier Inc. 15A57
1. Introduction Let d ∈ N and
Keywords: Jacobi matrix Eigenvalue Inverse problem α Submatrix ∈ (1, 2). We consider
the following pseudo-differential equation ut − ∆α/2 u + b · ∇ (u|u|q ) = 0, t > 0, x ∈ Rd , u(0, x) = M δ0 (x),
(1.1)
where M > 0 is arbitrary constant and b ∈ Rd is a constant vector. In this paper, we focus on the critical case q = (α − 1)/d. Here, ∆α/2 denotes the fractional Laplacian defined by the Fourier transform α/2 φ(ξ) = −|ξ|α φ(ξ), ∆
φ ∈ Cc∞ (Rd ).
Eq. (1.1) for various values of q and initial conditions u0 was recently intensely studied [2,4,3,7]. For d = 1, the case q = 2 is of particular interest (see e.g. [14,1,15,19]) because it is a natural counterpart of the classical Burgers equation. Another interesting value of q is α−1 d . In [4] authors proved that the solution of (1.1), which we denote throughout the paper by uM (t, x), exists and is unique and positive. It belongs also to E-mail address:
[email protected]. 1 +86 13914485239. [1, ∞]. The exponent q = α−1 Lp Rd for every p ∈ Tel.: d is critical in some sense. It is the only value for which the function uM (t, x) is self-similar. It satisfies the following scaling condition [4] http://dx.doi.org/10.1016/j.laa.2014.09.031 0024-3795/© 2014 Published by Elsevier Inc. uM (t, x) = ad uM (aα t, ax),
for all a > 0.
∗ Corresponding author. E-mail addresses:
[email protected] (T. Jakubowski),
[email protected] (G. Serafin).
http://dx.doi.org/10.1016/j.na.2015.10.016 0362-546X/© 2015 Elsevier Ltd. All rights reserved.
(1.2)
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Furthermore, the linear and nonlinear terms in (1.1) have equivalent influence on the asymptotic behavior of the solution. If q > (α − 1)/d, the operator ∆α/2 plays the main role. More precisely, for such q and a function u satisfying (1.1), with not necessarily the same initial condition, we have α/2 lim tn(1−1/p)/α u(t, ·) − e∆ u(0, ·) = 0, for each p ∈ [1, ∞]. t→∞
p
For q < (α − 1)/d another asymptotic behavior is expected. In addition, taking q = α−1 d for d = 1 and α = 2 we obtain the classical case, which makes Eq. (1.1) with critical exponent q one of the natural generalizations of the Burgers equation. Till the end of the paper we assume that d ≥ 1, α ∈ (1, 2) and q = α−1 d . Let p(t, x) be the fundamental solution of vt = ∆α/2 v.
(1.3)
In [7] the authors proved that for sufficiently small M there is a constant C = C(d, α, M, b) such that uM (t, x) ≤ Cp(t, x),
t > 0, x ∈ Rd .
(1.4)
In this paper we get rid of the smallness assumption of M . Furthermore, we also obtain the lower bounds of uM . We propose a new method which allows us to show pointwise estimates of solutions to the nonlinear problem (1.1) without the smallness assumption imposed on M . This method has been inspired by the proof of [6, Theorem 1]. Our main result is Theorem 1.1. Let d ≥ 1 and α ∈ (1, 2). Let uM (t, x) be the solution of Eq. (1.1) with q = a constant C = C(d, α, M, b) such that C −1 p(t, x) ≤ uM (t, x) ≤ Cp(t, x),
α−1 d .
There exists
t > 0, x ∈ Rd .
In addition, applying Theorem 1.1, we get the following estimates |uM (1, x) − M p(1, x)| ≤ c |∇uM (t, x)| ≤ c
p(1, x) , 1 + |x|
p(1, x) . + |x|
t−1/α
The fractional Laplacian plays also a very important role in the probability theory as a generator of the so called isotropic stable process. The theory of its linear perturbations has been recently significantly developed, see e.g., [5,6,12,13,18,16,17,8,10]. However, since the term b · ∇(|u|q u) in (1.1) represents a nonlinear drift, methods used in the linear case often cannot be adapted. In the proofs we mostly use the Duhamel formula and its suitable iteration. The scaling condition (1.2) is also intensively exploit. The paper is organized as follows. In Preliminaries we collect some properties of the function p(t, x) and introduce the Duhamel formula as well. In Section 3 we prove Theorem 1.1. In Section 4 we apply the methods and the results of Section 3 to obtain estimates of |uM (1, x) − M p(1, x)| and |∇uM (1, x)|. 2. Preliminaries 2.1. Notation For two positive functions f, g we denote f . g whenever there exists a constant c > 1 such that f (x) < cg(x) for every argument x. If f . g and g . f we write f ≈ g. If value of a constant in estimates is relevant, we denote it by Ck , k ∈ N, and it does not change throughout the paper.
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2.2. Properties of p(t, x) The fundamental solution of (1.3) may be given by the inverse Fourier transform α −d p(t, x) = (2π) e−ix·ξ e−t|ξ| dξ, t > 0, x ∈ Rd . Rd
This implies the following scaling property p(t, x) = ad p(aα t, ax),
for all a > 0.
(2.1)
Note that uM (t, x) possesses exactly the same property. Let p(t, x, y) := p(t, y − x). Below, we give two well-known estimates of p and the gradient of p (see [5] for more details). p(t, x, y) ≈
t
t1/α
|∇y p(t, x, y)| ≈
d+α , + |y − x| t1/α
t |y − x| d+2+α . + |y − x|
(2.2) (2.3)
Additionally, we denote t1/α
p(t, x, y) . + |x − y|
(2.4)
|∇y p(t, x, y)| . pˆ(t, x, y).
(2.5)
pˆ(t, x, y) = This implies
By (2.1), the function pˆ possesses a scaling property pˆ(t, x) = ad+1 pˆ(aα t, ax),
a > 0.
It satisfies also the following Chapman–Kolmogorov-like inequality. Lemma 2.1. For any t, s > 0 and x, y ∈ Rd , we have pˆ(t, x, z)ˆ p(s, z, y) dz ≤ C(t−1/α ∨ s−1/α )ˆ p(t + s, x, y).
(2.6)
Rd
Proof. We note that pˆ(t, x, y) is comparable with the stable density in dimension d + 1, so it satisfies the following 3P inequality (see [6]) pˆ(t, x, z)ˆ p(s, z, y) ≤ c pˆ(t + s, x, y)(ˆ p(t, x, z) + pˆ(s, z, y)). Hence,
pˆ(t, x, z)ˆ p(s, z, y) dz ≤ c pˆ(t + s, x, y)
Rd
(t−1/α p(t, x, z) + s−1/α p(s, z, y)) dz
Rd
≤ 2c (t−1/α ∨ s−1/α )ˆ p(t + s, x, y). In the sequel, we will need the below-given estimates of integrals involving the function p(t, x) as well as its gradient. We present them in two separate lemmas, since in the proof of the latter one more sophisticated methods are required.
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Lemma 2.2. For any 0 < γ < α/d, we have 1 |∇z p(1 − s, x, z)|p(s, z)1+γ dz ds ≤ c p(1, x)1+γ ∨ pˆ(1, x) . 0
Proof. For |x| < 1 we have 1 1+γ |∇z p(1 − s, x, z)|p(s, z) dz ds ≤ 0
1
0
Rd
(1 − s)−1/α s−γd/α p(1 − s, x, z)p(s, z) dz ds
Rd
≤ p(1, x) ≈ p(1, x)1+γ ≈ pˆ(1, x).
Furthermore, for |x| ≥ 1, 1 |∇z p(1 − s, x, z)|p(s, z)1+γ dz ds Rd 0 1
≤ c2
(2.7)
Rd
0
pˆ(1, x)p(s, z)sdγ/α dz ds +
1
0
|z|≤|x|/2
pˆ(1 − s, x, z)p(1, x)1+γ dz ds
|z|>|x|/2
≤ c3 (ˆ p(1, x) ∨ p(1, x)1+γ ). If we keep the absolute value outside the integrals, we can obtain better estimates. Lemma 2.3. For all x ∈ Rd , we have 1 1+q . pˆ(1, x), b · ∇ p(1 − s, z)p(s, x, z) dz ds z d 0
(2.8)
R
and 1/2 1 1+q q b · ∇z p(1 − s, x, z)∇z [p(s, z)] dz ds . pˆ(1, x) + p(1, x) . 0 1 + |x| Rd
(2.9)
Proof. Since p(1, x) ≈ 1 and pˆ(1, x) ≈ 1 for |x| ≤ 1, obviously (2.8) and (2.9) hold for |x| < 1. Hence we will consider only |x| ≥ 1. Note that for any z ∈ Rd and f ∈ C 1 (Rd ), we have f (x + z) − f (x) = z∇x f (x + rz),
(2.10)
for some r ∈ (0, 1). Hence, for z ∈ B(0, |x|/2), we obtain |p(s, x, z)1+q − p(s, x, −z)1+q | ≤ |p(s, x − z)1+q − p(s, x)1+q | + |p(s, x + z)1+q − p(s, x)1+q | ≤ c|z|p(s, x)q pˆ(s, x). Since ∇z p(s, z) = −∇z p(s, −z), we get 1 b · ∇z p(1 − s, z)p(s, x, z)1+q dz ds = − 0
B(0,|x|/2)
0
1
(2.11)
b · ∇z p(1 − s, z)p(s, x, −z)1+q dz ds.
B(0,|x|/2)
Consequently, by (2.11), 1 1+q b · ∇z p(1 − s, z)p(s, x, z) dz ds 0 B(0,|x|/2) 1 1 = b · ∇z p(1 − s, z)(p(s, x, z)1+q − p(s, x, −z)1+q )dz ds 2 0 B(0,|x|/2)
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1
.
0 1
.
pˆ(1 − s, z)|p(s, x, z)1+q − p(s, x, −z)1+q |dz ds
B(0,|x|/2)
p(1 − s, z) |z|p(s, x)q pˆ(s, x)dzds . |z|
0
B(0,|x|/2)
0
1
s(1−α)/α pˆ(s, x)ds . pˆ(1, x).
Therefore,
1
b · ∇z p(1 − s, z)p(s, x, z)1+q dzds
0
Rd 1
.
0
pˆ(1 − s, x)s(1−α)/α p(s, x, z)dzds + pˆ(1, x) . pˆ(1, x).
B(0,|x|/2)c
The proof of (2.9) is based on the same argument but for a convenience of the reader, we give the details. First, for |z| < |x|/2 and s ∈ (0, 1/2), (2.10) yields |b · ∇x p(1 − s, x, z) − b · ∇x p(1 − s, x, −z)| ≤ c|z|
p(1, x) . 1 + |x|2
Since 1/2
0
[b · ∇z p(1 − s, x, z)] ∇z [p(s, z)]1+q dz ds
B(0,|x|/2)
=−
0
1/2
[b · ∇z p(1 − s, x, −z)] ∇z [p(s, z)]1+q dz ds,
B(0,|x|/2)
we get 1/2 1+q [b · ∇z p(1 − s, x, z)] ∇z [p(s, z)] dz ds 0 d R 1/2 . |b · ∇x p(1 − s, x, z) − b · ∇x p(1 − s, x, −z)| |∇z [p(s, z)]1+q | dz ds 0
B(0,|x|/2)
+ .
1/2
0 1/2
pˆ(1 − s, x, z)|∇z [p(s, z)]1+q | dz ds
B(0,|x|/2)c
0
B(0,|x|/2)
. pˆ(1, x)
p(1, x) p(s, z) p(s, z)q dz ds + 2 1 + |x| |z| + p(1, x)q .
|z| 1 1 + |x|
0
1/2
pˆ(1 − s, x, z)p(1, x)q pˆ(1, x) dz ds
B(0,|x|/2)c
2.3. Duhamel formula Our main tool is the following Duhamel formula, t uM (t, x) = M p(t, x) + p(t − s, x, z)b · ∇z [uM (s, z)]1+q dz ds. 0
(2.12)
Rd
In the following, we assume that uM (t, x) = t−d/α uM (1, xt−1/α ) is a nonnegative self-similar solution of Eq. (2.12) such that uM (1, ·) ∈ Lp Rd for each p ∈ [1, ∞]. As it was mentioned in Introduction, the existence of such a function was shown in [4]. It turns out that the integral in (2.12) is not absolutely convergent, but integrating by parts we obtain a more convenient form t uM (t, x) = M p(t, x) − b · ∇z p(t − s, x, z)[uM (s, z)]1+q dz ds, (2.13) 0
Rd
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which is absolutely convergent, cf. (2.7). Indeed, we have ∥uM (s, ·)∥p . s−d(1−1/p)/α , p ∈ [1, ∞] (see [4]). Hence, by (2.2) and (2.3), t b · ∇z p(t − s, x, z)[uM (s, z)]1+q dz ds 0
Rd
.
1+q
|∇z p(t − s, x, z)|[uM (s, z)]
0
.
t/2
0
t/2
(t − s)−(d+1)/α [uM (s, z)]1+q dz ds +
Rd
. t−(d+1)/α
t
t/2
0
|∇z p(t − s, x, z)|[uM (s, z)]1+q dz ds
dz ds +
Rd t/2
t
t/2
s−(α−1)/α ds + t−d(1+q)/α
Rd
(t − s)−1/α p(t − s, x, z)s−d(1+q)/α dz ds
Rd
t
(t − s)−1/α ds . t−d/α .
t/2
Now, using the scaling property (1.2), we obtain t b · ∇z p(t − s, x, z)[uM (s, z)]1+q dz ds 0 Rd t = b · ∇z p(t − s, x, z)s−(d−1+α)/α [uM (1, s−1/α z)]1+q dz ds 0 Rd t = s(1−α)/α b · ∇w p(t − s, x, s1/α w)[uM (1, w)]1+q dw ds 0
Rd
=α
0
t1/α
b · ∇w p(t − rα , x, rw)[uM (1, w)]1+q dw dr.
(2.14)
Rd
Finally, we get uM (t, x) = M p(t, x) − α
t1/α
0
∇w p(t − rα , x, rw)[uM (1, w)]1+q dw dr.
(2.15)
Rd
Due to the scaling property (1.2) it suffices to consider uM (t, x) only for t = 1. By (2.5) and (2.15), there exists a constant C1 > 0 such that for every x ∈ Rd , 1 uM (1, x) ≤ M p(1, x) + C1 pˆ(1 − rα , x, rw)[uM (1, w)]1+q dw dr. (2.16) 0
Rd
3. Proof of Theorem 1.1 In this section, we prove the main theorem of the paper. First, we will show that the function uM (1, x) vanishes at infinity. Lemma 3.1. Let f ∈ C 1 (Rd ) ∩ L1 (Rd ) be a nonnegative function such that ∇f ∈ L∞ . Then, lim f (x) = 0.
|x|→∞
Proof. Fix ε > 0. Let K = ∥∇f ∥∞ and R > 0. Since f ∈ C 1 , we have f (w) ≥ f (x) − K|w − x0 | for any x, w ∈ Rd . Thus, for any x > R + ε/K, |B(0, 1)|εd (f (x) − ε). fR := f (w)dw ≥ (f (x) − K|w − x|) dw > Kd |w|>R |w−x|<ε/K Since fR → 0 as R → ∞, f (x) < 2ε for |x| > R and sufficiently large R. This ends the proof.
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Since uM (1, ·) ∈ C ∞ (Rd ) and ∇uM (1, ·) ∈ L∞ (Rd ) (see [9,4]), by Lemma 3.1, we immediately get Proposition 3.2. We have lim|x|→∞ uM (1, x) = 0. Now, we introduce a series of auxiliary functions. We put 1 H(x, w) = pˆ(1 − rα , x, rw)dr, 0
(3.1)
and, for β ∈ (0, 1), ˜ H(x, w) =
1
0
r−β pˆ(1 − rα , x, rw)dr,
where the function pˆ(t, x) is given by (2.4). Additionally, for R > 0 we define ˜ ˜ H(x, w)[uM (1, w)]1+q dw, hR (x) =
(3.2)
(3.3)
B(0,R)
˜ H(x, w)uM (1, w)dw.
(3.4)
H(x, w)p(1, w)dw ≤ C2 p(1, x),
(3.5)
H(x, w)ˆ p(1, w)dw ≤ C3 pˆ(1, x).
(3.6)
˜ R (x) = H
B(0,R)c
˜ Note that H(x, w) ≤ H(x, w). Lemma 3.3. For x ∈ Rd , we have Rd
Rd
Proof. By scaling property and Chapman–Kolmogorov equation for the function p(t, x, y), we get p(1 − sα , x, sw)p(1, w) dw = s−d p(s−α − 1, s−1 x, w)p(1, w) dw Rd
Rd
= s−d p(s−α , s−1 x) = p(1, x). Consequently,
1
H(x, w)p(1, w)dw ≤
0
Rd
(1 − sα )−1/α p(1 − sα , x, sw)p(1, w)dw ds
Rd
1
= p(1, x)
0
(1 − sα )−1/α ds,
which proves (3.5). Furthermore, scaling property and (2.6) give us 1 H(x, w)ˆ p(1, w)dw = s−(d+1) pˆ(s−α − 1, s−1 x, w)ˆ p(1, w) dw ds 0
Rd
≤c
0
1
Rd
s−(d+1) ((s−α − 1)−1/α ∨ 1)ˆ p(s−α , s−1 x)ds
= c pˆ(1, x)
0
1
(s(1 − sα )−1/α ∨ 1)ds = C3 pˆ(1, x).
The next step is to provide a Chapman–Kolmogorov-like inequality involving functions H(x, w) and ˜ H(x, w). At first, we present a technical lemma.
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Lemma 3.4. Let β > 0 be fixed. There is a constant C such that for v ∈ (0, 1), we have 1 r−(1+β) [r(rα − v α )−1/α + (1 − rα )−1/α ]dr ≤ Cv −β .
(3.7)
v
Proof. First, substituting r = sv, we estimate 1/v 1 s−1−β (sα − 1)−1/α ds r−(1+β) v(rα − v α )−1/α dr = v −β v 1 ∞ −β s−1−β (sα − 1)−1/α ds = c1 v −β . ≤v 1
Hence, it suffices to prove 1
r−(1+β) (1 − rα )−1/α dr ≤ c2 v −β .
v
For v ≥ 1/2, it is obvious. For v < 1/2, we estimate 1 r−(1+β) (1 − rα )−1/α dr ≤ c3 v
1/2
r−(1+β) dr +
v
1
1/2
(1 − rα )−1/α dr r1+β
v −β ≤ c3 + c4 ≤ c2 v −β . β This allows us to prove the following lemma. Lemma 3.5. There exists a constant C4 > 0 such that for x, z ∈ Rd , we have ˜ ˜ H(x, w)H(w, z)dw ≤ C4 H(x, z).
(3.8)
Rd
Proof. Scaling property and Lemma 2.1 give us 1 1 ˜ H(x, w)H(w, z)dw = r−β pˆ(1 − sα , x, sw)ˆ p(1 − rα , w, rz)dw ds dr 0
Rd
=
0
1
0
0
= c1
r−β s−(d+1) pˆ(s−α − 1, s−1 x, w)ˆ p(1 − rα , w, rz)dw ds dr
0 1 1
Rd
≤ c1
Rd
1
0
1
0
r−β [(s−α − 1)−1/α + (1 − rα )−1/α ]s−(d+1) pˆ(s−α − rα , s−1 x, rz)ds dr
1
r−β [s(1 − sα )−1/α + (1 − rα )−1/α ]ˆ p(1 − sα rα , x, srz)ds dr.
0
Substituting s = v/r in the inner integral and then using Fubini–Tonelli theorem and (3.7), we get 1 1 ˜ H(x, w)H(w, z)dw ≤ c1 pˆ(1 − v α , x, vz) r−(1+β) [v(rα − v α )−1/α + (1 − rα )−1/α ]dr dv Rd
0
v
˜ ≤ c2 H(x, w). As a consequence, we get Corollary 3.6. For x ∈ Rd , we have
˜ R (w)dw ≤ C4 h ˜ R (x), H(x, w)h
(3.9)
˜ R (w)dw ≤ C4 H ˜ R (x). H(x, w)H
(3.10)
Rd
Rd
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Now, we pass to the proof of the main result of the paper. Proof of Theorem 1.1. Let C0 = C1 (C2 ∨ C4 ). By Proposition 3.2, we may choose η ∈ (0, 1) and R > 0 such 1/q for |x| > R. Thus, by (2.16), we have that |uM (1, x)| < Cη0 C1 η ˜ uM (1, x) ≤ M p(1, x) + C1 hR (x) + H(x, w)uM (1, w)dw (3.11) C0 B(0,R)c C1 η ˜ HR (x). C0
˜ R (x) + ≤ M p(1, x) + C1 h
(3.12)
We put (3.12) to (3.11) and, by Lemma 3.3 and Corollary 3.6, we have ˜ R (x) uM (1, x) ≤ M p(1, x) + C1 h C1 η C1 η ˜ ˜ HR (w) dw + H(x, w) M p(1, w) + C1 hR (w) + C0 B(0,R)c C0 ˜ R (x) + η 2 H ˜ R (x). ≤ M (1 + η)p(1, x) + C1 (1 + η)h
(3.13)
(3.14)
Now, we put (3.14) into (3.11) and, by Lemma 3.3 and Corollary 3.6, we obtain ˜ R (x) + η 3 H ˜ R (x). uM (1, x) ≤ M (1 + η + η 2 )p(1, x) + C1 (1 + η + η 2 )h Hence, by induction, uM (1, x) ≤ M
n
n
η k p(1, x) + C1
k=0
˜ R (x) + η n+1 H ˜ R (x). ηk h
k=0
Taking n → ∞, we get uM (1, x) ≤
C1 ˜ M p(1, x) + hR (x). 1−η 1−η
(3.15)
Furthermore, since both functions p(1, ·) and uM (1, ·) are continuous and positive (see [4, Theorem 2.1]), they are comparable on every compact set. Hence, we focus only on large values of |x|. We first prove the upper estimate. Let |x| > 2R. For |w| < R and s ∈ (0, 1), we have |x − sw| > |x|/2, and consequently pˆ(s, x, sw) ≈ |x|−d−α−1 . p(1, x). Hence, 1 ˜ R (x) = h r−β pˆ(1 − rα , x, rw)[uM (1, w)]1+q dr dw B(0,R)
.
0
1+q p(1, x) ∥uM (1, ·)∥∞
B(0,R)
0
1
r−β dr dw
1+q
≈ ∥uM (1, ·)∥∞ p(1, x), which, by (3.15), gives us uM (1, x) . p(1, x).
(3.16)
To provide lower estimates we show that the integral in (2.15) is much smaller than M p(t, x) for sufficiently large |x|. Using (3.16), (2.7) and scaling property, we get 1 α 1+q α b · ∇ p(1 − s , x, sw)[u (1, w)] dw ds w M 0
≤α
Rd 1
0
|b · ∇w p(1 − sα , x, sw)| [p(1, w)]1+q dw ds
Rd
≤ c1 pˆ(1, x) ∨ p(1, x)1+q ≤ c2 p(1, x) |x|−1 + |x|−q(d+α) ,
(3.17)
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405
which implies uM (1, x) ≥ ˜ > 0. The proof is complete. for some R
M p(t, x), 2
˜ |x| > R,
4. Asymptotics and gradient estimates In this section we provide asymptotics of the integral in the Duhamel formula, i.e. of the expression uM (1, x)−M p(1, x), as well as the upper bound of the gradient of uM (t, x). We collect those results together, since they turn out to be related and proofs of both of them base on some common ideas. In the proof of Theorem 1.1 (see (3.17)), we showed that |uM (1, x) − M p(1, x)| . p(1, x)1+q ∨ pˆ(1, x).
(4.1)
The next lemma improves that result. Proposition 4.1. There is a constant C > 0 such that |uM (1, x) − M p(1, x)| ≤ C
p(1, x) , 1 + |x|
x ∈ Rd .
Proof. We note that it suffices to consider only |x| > 1. If p(1, x)1+q . pˆ(1, x) (i.e. q ≥ 1/(d + α)), we note that the assertion of the proposition holds. Suppose q < 1/(d + α). Note that for 0 < v < c u, |v 1+q − u1+q | . (v q + uq )|v − u| ≤ (cq + 1)uq |v − u|.
(4.2)
Hence, by (2.8), (2.7), (4.1) and scaling properties of the functions uM (t, x) and p(t, x), we get 1 b · ∇z p(1 − s, x, z)[(M p(s, z))1+q + uM (s, z)1+q − (M p(s, z))1+q ] dz ds |uM (1, x) − M p(1, x)| = 0 Rd 1 1+q b · ∇z p(1 − s, x, z)[p(s, z)] dz ds . 0
+
0
Rd 1
|b · ∇z p(1 − s, x, z)|p(s, z)q s−d/α |uM (1, zs−1/α ) − M p(1, zs−1/α )| dz ds
Rd
. pˆ(1, x) +
0
1
|∇z p(1 − s, x, z)|sqd/α [p(s, z)]1+2q dz ds . pˆ(1, x) ∨ p(1, x)1+2q .
Rd
By induction, for sufficiently large n, we get |uM (1, x) − M p(1, x)| . pˆ(1, x) ∨ p(1, x)1+nq . pˆ(1, x). Finally, we give estimates of |∇x uM (t, x)|, cf. (2.5). Theorem 4.2. There exists a constant C > 0 such that |∇x uM (t, x)| ≤ C for all t > 0 and x ∈ Rd .
t , (|x| + t1/α )d+α+1
406
T. Jakubowski, G. Serafin / Nonlinear Analysis 130 (2016) 396–407
Proof. By scaling property ∇x uM (x, t) = t−(d+1)/α uM (1, xt−1/α ), it is enough to consider t = 1. Using (2.13) and integrating by parts, we obtain 1/2
∇x uM (1, x) = M ∇x p(1, x) − +
1
1/2
0
∇z [b · ∇z p(1 − s, x, z)] [uM (s, z)]1+q dz ds
Rd
b · ∇z p(1 − s, x, z)∇z [uM (s, z)]1+q dz ds.
(4.3)
Rd
Note that |∇z b · ∇z p(1 − s, x, z)| . pˆ(1, x, z) for s ∈ (0, 1/2) (see, e.g., [11]). We estimate the first integral in (4.3) employing Proposition 4.1, integration by parts and the formula (4.2) as follows 1/2 1+q ∇z [b · ∇z p(1 − s, x, z)] [uM (s, z)] dz ds 0 Rd 1/2 1+q 1+q 1+q = ∇z [b · ∇z p(1 − s, x, z)] [(M p(s, z)) + uM (s, z) − (M p(s, z)) ] dz ds 0 d R 1/2 . [b · ∇z p(1 − s, x, z)] ∇z [p(s, z)]1+q dz ds 0 Rd 1/2 + pˆ(1, x, z)[p(s, z)]q |uM (s, z) − M p(s, z)| dz ds. (4.4) 0
Rd
By Proposition 4.1 and scaling property, [p(s, z)]q |uM (s, z) − M p(s, z)| . p(s, z)q s1/α pˆ(s, z) . s2/α−1 pˆ(s, z). Hence, using formulae (4.4), (2.9) and (2.6), we get 1/2 1+q ∇z [b · ∇z p(1 − s, x, z)] [uM (s, z)] dz ds 0 d R 1/2 1 . pˆ(1, x) + p(1, x)q + s1/α−1 pˆ(1, x) ds . pˆ(1, x). 1 + |x| 0 Applying this to (4.3) and repeating operations from (2.14), we arrive at |∇x uM (1, x)| . pˆ(1, x) + . pˆ(1, x) +
1
−1
s
1/2 1 0
|b · ∇z p(1 − sα , x, sz)|[uM (1, z)]q |∇z uM (1, z)| dz ds
Rd
pˆ(1 − sα , x, sz)[p(1, z)]q |∇z uM (1, z)| dz ds.
Rd
Now, we are going to adapt the method developed in the previous section. For R > 0, we define ˜ g˜R (x) = H(x, w)[p(1, w)]q |∇w uM (1, w)|dw, B(0,R) ˜ R (x) = ˜ G H(x, w)|∇w uM (1, w)|dw, B(0,R)c
˜ where H(x, w) and H(x, w) are given by (3.1) and (3.2), respectively. Thus |∇x uM (1, x)| ≤ c1 pˆ(1, x) + g˜R (x) + H(x, w)[p(1, w)]q |∇w uM (1, w)| dw , B(0,R)c
T. Jakubowski, G. Serafin / Nonlinear Analysis 130 (2016) 396–407
407
for some constant c1 > 0. Let c2 = c1 (C3 ∨ C4 ). We may choose η ∈ (0, 1) and R > 0 such that 1/q for |x| > R. Thus, |p(1, x)| < cη2 c1 η |∇x uM (1, x)| ≤ c1 pˆ(1, x) + c1 g˜R (x) + H(x, w)|∇z uM (1, w)|dw c2 B(0,R)c c1 η ˜ ≤ c1 pˆ(1, x) + c1 g˜R (x) + GR (x). c2 By Lemma 3.5, we have H(x, w)˜ gR (w)dw ≤ C4 g˜R (x), Rd
˜ R (w)dw ≤ C4 G ˜ R (x). H(x, w)G
Rd
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