Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 199, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
FRACTIONAL POROUS MEDIUM AND MEAN FIELD
EQUATIONS IN BESOV SPACES
XUHUAN ZHOU, WEILIANG XIAO, JIECHENG CHEN
Abstract. In this article, we consider the evolution model
∂t u − ∇ · (u∇P u) = 0,
P u = (−∆)−s u,
0 < s ≤ 1, x ∈ Rd , t > 0.
We show that when s ∈ [1/2, 1), α > d + 1, d ≥ 2, the equation has a unique
α . Moreover, in the critical
local in time solution for any initial data in B1,∞
α , 2 ≤ p ≤ ∞, α > d/p, d ≥ 3.
case s = 1, the solution exists in Bp,∞
1. Introduction
Let 0 < s ≤ 1, we consider the evolution equation
∂t u − ∇ · (u∇P u) = 0,
P u = (−∆)−s u,
u(x, 0) = u0 (x).
(1.1)
d
where x ∈ R , t > 0. u = u(x, t) is a real-valued function, representing the density
or concentration. P represents the pressure.
When 0 < s < 1, we refer the equation as a fractional porous medium equation.
It was first introduced by Caffarelli and Vázquez [2]. They proved the existence
of a weak solution when u0 is a bounded function and has exponential decay at
infinity. Caffarelli, Soria, and Vázquez [1] studied its regularity theory of the weak
solution with u0 ∈ L1 ∩ L∞ and the continuity of bounded solutions.
When s = 1, the equation leads to a mean field equation
ut = ∇ · (u∇P u),
P u = (−∆)−1 u,
u(x, 0) = u0 (x).
(1.2)
Equation that was first studied by Lin and Zhang [7]. They proved the existence
and uniqueness of positive L∞ solution in two dimensions. When d ≥ 3, Vázquez
and Serfaty [11] studied the existence and uniqueness of the weak solution in L∞
spaces for (1.2) by taking the limit s → 1 in the weak solutions of (1.1). Since
papers in literature only addressed the weak solution of (1.2), in this article we
α
show that the strong solution exists in Bp,∞
, α > 0.
In this article, we are interested in finding the strong solutions of (1.1) in the
α
Besov spaces Bp,∞
. We will show that when d ≥ 2, s ∈ [1/2, 1), α > d + 1,
α
equation (1.1) has a unique local in time solution for any initial data in B1,∞
.
α
When s = 1, the solution extends to Bp,∞ , α > d/p, 2 ≤ p ≤ ∞, d ≥ 3. The idea
2000 Mathematics Subject Classification. 35K55, 35K65, 76S05.
Key words and phrases. Fractional porous medium equation; mean field equation;
local solution; Besov space.
c
2014
Texas State University - San Marcos.
Submitted April 10, 2014. Published September 23, 2014.
1
2
X. ZHOU, W. XIAO, J. CHEN
EJDE-2014/199
of our proof is inspired by the methods used in [3, 12], where the authors studied
the quasi-geostrophic equation. In our proof, we construct two commutators, give
three estimates of them, and construct a function sequence fitting our equation.
The rest of this article is divided in four parts. Section 2 recalls the definition and
some properties of the Besov spaces, the Bernstein inequality involving both integer
and fractional derivatives, as well as some properties of the fractional Laplacian.
In Section 3 we prove three estimates about the constructed commutators and a
priori estimate of the solution. Section 4 proves the existence and uniqueness of the
fractional porous medium equation. Section 5 is devoted to the mean field equation.
The main results are the following two theorems.
Theorem 1.1. Let d ≥ 2, s ∈ [1/2, 1], α > d + 1. Assume that the initial data u0 ∈
α
α
B1,∞
. Then we can find T = T (ku0 kB1,∞
), such that a unique solution u to (1.1) on
β
α+2s−2
[0, T ]×Rd exists. And the solution belongs to C 1 ([0, T ]; B1,∞
)∩L∞ ([0, T ]; B1,∞
),
and β ∈ [α + 2s − 2, α].
Theorem 1.2. Let d ≥ 3, α > d/p when 2 ≤ p ≤ ∞. Assume that the initial data
α
α
u0 ∈ Bp,∞
. Then we can find T = T (ku0 kBp,∞
), such that a unique solution u to
the given mean field equation (1.2) on [0, T ] × Rd exists. And the solution belongs
α
β
), β ∈ (d/p, α), 2 ≤ p ≤ ∞.
) ∩ L∞ ([0, T ]; Bp,∞
to C 1 ([0, T ]; Bp,∞
2. Preliminaries
In this section, we recall the definition of the Besov space. We start with a
dyadic decomposition of Rd .
Suppose χ ∈ C0∞ (Rd ), φ ∈ C0∞ (Rd \ {0}) satisfying
4
},
3
3
8
supp φ ⊂ {ξ ∈ Rd : < |ξ| < },
4
3
X
χ(ξ) +
φ(2−j ξ) = 1, ξ ∈ Rd ,
supp χ ⊂ {ξ ∈ Rd : |ξ| ≤
j≥0
X
φ(2−j ξ) = 1,
ξ ∈ Rd \{0}.
j∈Z
Define the operators
−j
Z
h(2j y)u(x − y) dy,
Z
X
Sj f =
∆k f = χ(2−j D)u = 2jd g(2j y)u(x − y) dy,
∆j u = φ(2
jd
D)u = 2
(2.1)
k≤j−1
∨
where g = χ and h = φ∨ are the inverse Fourier transform of χ and φ, respectively.
It can be easily verified that with our choice of φ,
∆j ∆k f ≡ 0, if |j − k| ≥ 2,
∆j (Sk−1 f ∆k f ) ≡ 0,
if |j − k| ≥ 5.
(2.2)
Definition 2.1. For any α ∈ R, and p, q ∈ [1, ∞], the homogeneous Besov spaces
r
Ḃp,q
are defined as
α
Ḃp,q
= {f ∈ Z 0 (Rd ) : kf kḂ α < ∞}.
p,q
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FRACTIONAL POROUS MEDIUM EQUATION
3
Here,
kf kḂ α =
hX
p,q
2jαq k∆j f kqLp
i1/q
,
where q < ∞.
j∈Z
When q = ∞,
kf kḂ α
p,∞
= sup 2jα k∆j f kLp .
j∈Z
Z (R ) denotes the dual space of Z(R ) = {f ∈ S(Rd ) : ∂ γ fˆ(0) = 0, ∀γ ∈ Nd } and
can be identified by the quotient space of S 0 /P with the polynomials space P.
0
d
d
Definition 2.2. For any α ∈ R, and p, q ∈ [1, ∞], the inhomogeneous Besov space
r
Bp,q
is defined as
α
α
Bp,q
= {f ∈ S 0 (Rd ) : kf kBp,q
< ∞}.
Here,
α
kf kBp,q
=
∞
X
2jαq k∆j f kqLp
1/q
+ kS0 (f )kLp ,
when q < ∞.
j≥0
When q = ∞,
α
kf kBp,∞
= sup 2jα k∆j f kLp + kS0 (f )kLp .
j≥0
Let us state some basic properties of the Besov spaces.
Proposition 2.3. Let s ∈ R, 1 ≤ p ≤ ∞, 1 ≤ q ≤ ∞.
α
α
α
∩ Lp , and kf kBp,q
= kf kḂ α + kf kLp ;
= Ḃp,q
(i) If α > 0, then Bp,q
p,q
α2
α1
α
α
(ii) If α1 ≤ α2 , then Bp,q
⊂ Bp,q
. If 1 ≤ q1 ≤ q2 ≤ ∞, then Ḃp,q
⊂ Ḃp,q
and
1
2
α
α
Bp,q1 ⊂ Bp,q2 ;
α
(iii) If α > dp , then Bp,q
,→ L∞ . If p1 ≤ p2 , α1 − pd1 > α2 − pd2 , then Bpα11,q1 ,→
α
α
α2
;
Bp2 ,q2 , Bp,min(p,2) ,→ Hpα ,→ Bp,max(p,2)
α
α
α
(iv) If α > 0, p ≥ 1, then kuvkBp,∞
≤ CkukL∞ kvkBp,∞
+ CkukBp,∞
kvkL∞ .
We now turn to Bernstein’s inequalities. When the Fourier transform of a function is supported on a ball or an annulus, the Lp -norms of the derivatives of the
function can be bounded in terms of the Lp -norms of the function itself. And it also
exists when one replaces the derivatives by the fractional derivatives (see [6, 14]).
Proposition 2.4. Let 1 ≤ p ≤ q ≤ ∞, γ ∈ Nd . (1) If α ≥ 0 and supp fˆ ⊂ {ξ ∈
Rd : |ξ| ≤ K2j } for some K > 0 and integer j, then
1
1
k(−∆)α Dγ f kLq ≤ C2j(2α+|γ|)+jd( p − q ) kf kLp .
(2) If α ∈ R and supp fˆ ⊂ {ξ ∈ Rd : K1 2j ≤ |ξ| ≤ K2 2j } for some K1 , K2 > 0
and integer j, then
1
1
e j(2α+|γ|)+jd( p1 − q1 ) kf kLp ,
C2j(2α+|γ|)+jd( p − q ) kf kLp ≤ k(−∆)α Dγ f kLq ≤ C2
e are positive constants independent of j.
where C and C
Next we state two pointwise inequalities which were proved in [5, 13].
Proposition 2.5. Let 0 ≤ α ≤ 1, f ∈ C 2 (Rd ) decay sufficiently fast at infinity.
Then for any x ∈ Rd ,
2f (x)(−∆)α f (x) ≥ (−∆)α f 2 (x).
4
X. ZHOU, W. XIAO, J. CHEN
EJDE-2014/199
Proposition 2.6. Let 0 ≤ α ≤ 1, p1 = kl11 ≥ 0, p2 = kl22 ≥ 1 be rational numbers with l1 , l2 odd, and k1 l1 + k2 l2 even. Then for any f ∈ C 2 (Rd ) that decays
sufficiently fast at infinity, and for any x ∈ Rd ,
(p1 + p2 )f p1 (x)(−∆)α f p2 (x) ≥ p2 (−∆)α f p1 +p2 (x).
3. A priori estimate
Proposition 3.1. Let α > 0, s ∈ (0, 1), p ∈ [1, ∞] be given. Assume r > d/p.
Then there exists some constant C such that
r+1 kuk α+1−2s + Ckuk r+2−2s kvkB α
2jα k[∆j , ∂i (−∆)−s u]∂i vkLp ≤ CkvkBp,∞
, (3.1)
Bp,∞
Bp,∞
p,∞
where the brackets [, ] represents the commutator, namely
[∆j , ∂i (−∆)−s u]∂i v = ∆j (∂i (−∆)−s u)∂i v) − ∂i (−∆)−s u∆j (∂i v).
Proof. Using Bony’s para-product decomposition, we have
[∆j , ∂i (−∆)−s u]∂i v = L1 + L2 + L3 + L4 + L5 ,
where
X
L1 =
∆j [Sk−1 (∂i (−∆)−s u)∆k (∂i v)] − Sk−1 (∂i (−∆)−s u)∆k (∆j (∂i v)),
|k−j|≤4
X
L2 =
∆j [Sk−1 (∂i v)∆k (∂i (−∆)−s u)],
|k−j|≤4
X
L3 =
e k (∂i v)),
∆j (∆k (∂i (−∆)−s u)∆
k≥j−2
L4 =
X
Sk−1 (∆j (∂i v))∆k (∂i (−∆)−s u),
k
X
L5 =
∆j 0 (∆j (∂i v))∆j 00 (∂i (−∆)−s u).
|j 0 −j 00 |≤1
We shall estimate the above terms separately. First observe
Z
X
L1 =
2jd h(2j (x − y)) Sk−1 (∂i (−∆)−s u)(y)
|k−j|≤4
− Sk−1 (∂i (−∆)−s u)(x) ∆k (∆j (∂i v))(y) dy.
By Young’s inequality and Bernstein’s inequality,
Z
X
−j
−s
kL1 kLp ≤ C
2 k∇∂i (−∆) ukL∞ k∆j (∂i v)kLp |y||h(y)| dy
|k−j|≤4
≤ C2−j 2j k(−∆)1−s ukL∞ k∆j vkLp
r+2−2s .
≤ Ck∆j vkLp kukBp,∞
Similarly,
r+1 kuk α+1−2s .
kL2 kLp ≤ C2j(1−2s) k∇vkL∞ k∆j ukLp ≤ C2−jα kvkBp,∞
Bp,∞
We can also estimate,
kL3 kLp ≤ C
X
k≥j−2
k∆k (∂i (−∆)−s u)kLp k∇vkL∞
EJDE-2014/199
FRACTIONAL POROUS MEDIUM EQUATION
≤ C2−jα
X
5
r+1
2(j−k)α 2k(α+1−2s) k∆k ukLp kvkBp,∞
k≥j−2
−jα
≤ C2
α+1−2s kvk r+1 .
kukBp,∞
Bp,∞
To estimate L4 , by the definition of Sj and ∆j , we can observe that only k satisfying
k ≥ j survive. Thus
X
kL4 kLp ≤
Ck∇vkL∞ 2k(1−2s) k∆k ukLp
k≥j
≤ C2−jα
X
r+1 kuk α+1−2s
2(j−k)α kvkBp,∞
Bp,∞
k≥j
≤ C2
−jα
r+1 kuk α+1−2s .
kvkBp,∞
Bp,∞
Since ∆k ∆j = 0, for |j − k| ≥ 2, we have
r+1 kuk α+1−2s .
kL5 kLp ≤ Ck∇vkL∞ 2j(1−2s) k∆j ukLp ≤ C2−jα kvkBp,∞
Bp,∞
Collecting the estimates above, we obtain
2jα k[∆j , ∂i (−∆)−s u]∂i vkLp
j(α+1−2s)
r+1 k∆j ukLp + kuk α+1−2s kvk r+1
r+2−2s + 2
kvkBp,∞
≤ C2jα k∆j vkLp kukBp,∞
Bp,∞
Bp,∞
α
r+2−2s + kuk α+1−2s kvk r+1 .
≤ kvkBp,∞
kukBp,∞
Bp,∞
Bp,∞
This completes the proof.
Proposition 3.2. Let α, s, p, r be as in Proposition 3.1. Then there exists a constant C such that
r
α+2−2s + Ckuk r+2−2s kvkB α
2jα k[∆j , ∂i (−∆)−s u]∂i vkLp ≤ CkvkBp,∞
kukBp,∞
. (3.2)
Bp,∞
p,∞
When p = ∞, this inequality becomes: for any r > 0,
r
α+2−2s + Ckuk r+2−2s kvkB α
2jα k[∆j , ∂i (−∆)−s u]∂i vkL∞ ≤ CkvkB∞,∞
kukB∞,∞
.
B∞,∞
∞,∞
(3.3)
Proof. We want to give a new estimate of the commutator in Proposition 3.1.
Following the above proof, the estimate of L1 unchanged, we give different bounds
for L2 , L3 , L4 , L5 . First,
Z
X
L2 =
2jd h(2j (x − y))(Sk−1 ∂i v)(y)∆k (∂i (−∆)−s u)(y) dy
|k−j|≤4
X
=
2jd
Z
∂i h(2j (x − y))2j (Sk−1 v)(y)∆k (∂i (−∆)−s u)(y) dy
|k−j|≤4
−
X
2jd
Z
h(2j (x − y))(Sk−1 v)(y)∆k (∂ii (−∆)−s u)(y) dy.
|k−j|≤4
So we obtain
kL2 kLp ≤ C22−2s (k∂i hkL1 + khkL1 )kvkL∞ k∆j ukLp
α+2−2s kvkB r
≤ C2−jα kukBp,∞
.
p,∞
6
X. ZHOU, W. XIAO, J. CHEN
EJDE-2014/199
Similarly,
X
kL3 kLp ≤ C
2jd
Z
∂i h(2j (x − y))2j ∆k (∂i (−∆)−s u)(y)∆k v(y) dy
k≥j−2
− 2jd
Z
h(2j (x − y))∆k (∂ii (−∆)−s u)(y)∆k v(y) dy.
Hence we obtain,
X
kL3 kLp ≤ C
(2(j−k)(α+1) + 2(j−k)α )2k(α+2−2s) kvkL∞ k∆k ukLp
k≥j−2
α+2−2s kvkB r
≤ C2−jα kukBp,∞
.
p,∞
Also,
kL4 kLp ≤
X
C2j k∆j vkL∞ 2k(1−2s) k∆k ukLp
k≥j
≤ C2−jα
X
r
2(j−k)(α+1) 2k(α+2−2s) kvkBp,∞
k∆k ukLp
k≥j
−jα
≤ C2
r
α+2−2s .
kvkBp,∞
kukBp,∞
Finally,
r
α+2−2s .
kL5 kLp ≤ Ck∆j vkL∞ 2j(2−2s) k∆j ukLp ≤ C2−jα kvkBp,∞
kukBp,∞
Collecting the estimates above, we can obtain
α+2−2s
r+2−2s + kvkB r
2jα k[∆j , ∂i (−∆)−s u]∂i vkLp ≤ C2jα k∆j vkLp kukBp,∞
kukBp,∞
p,∞
α
r+2−2s + kuk α+2−2s kvkB r
.
≤ kvkBp,∞
kukBp,∞
Bp,∞
p,∞
This completes the proof.
Proposition 3.3. Let α > 0, s ∈ (0, 1), p ∈ [1, ∞]. Assume r >
exists a constant C such that
d
p.
Then there
r+1 kuk α+1−2s + Ckuk r+2−2s kvkB α
2jα k[∆j , v](−∆)1−s ukLp ≤ CkvkBp,∞
,
Bp,∞
Bp,∞
p,∞
(3.4)
where the brackets [, ] represents the commutator,
[∆j , v](−∆)1−s u = ∆j (v(−∆)1−s u) − v∆j ((−∆)1−s u).
(3.5)
Proof. This proposition is proved similarly to Proposition 3.1. We start the proof
by writing
[∆j , v](−∆)1−s u = I1 + I2 + I3 + I4 + I5 ,
where
X
I1 =
∆j (Sk−1 v∆k (−∆)1−s u) − Sk−1 v∆k (∆j (−∆)1−s u),
|k−j|≤4
X
I2 =
∆j (Sk−1 (−∆)1−s u)∆k v),
|k−j|≤4
X
I3 =
e k (−∆)1−s u),
∆j (∆k v)∆
k≥j−2
I4 =
X
k
Sk−1 (∆j (−∆)1−s u)∆k v,
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FRACTIONAL POROUS MEDIUM EQUATION
X
I5 =
7
∆j 0 v∆j 00 (∆k (−∆)1−s u).
|j 0 −j 00 |≤1
Similar to the proof for Proposition 3.1, first we observe that
Z
X
kI1 kLp = k
2jd h(2j (x − y))(Sk−1 v(y) − Sk−1 v(x))∆k (−∆)1−s u)(y)dykLp
|k−j|≤4
≤C
X
2−j k∇vkL∞ k∆j (−∆)1−s ukLp
Z
|y||h(y)| dy
|k−j|≤4
≤ C2−j 22j(1−s) k∇vkL∞ k∆j ukLp
r+1 .
≤ C2j(1−2s) k∆j ukLp kvkBp,∞
Also we obtain
r+2−2s k∆j vkLp ,
kI2 kLp ≤ Ck(−∆)1−s ukL∞ k∆j vkLp ≤ CkukBp,∞
X
kI3 kLp ≤ C
k∆k vkLp 2k(2−2s) k∆k ukL∞
k≥j−2
X
≤ C2−jα
2(j−k)α 2kα k∆k vkLp 2k(2−2s) k∆k ukL∞
k≥j−2
α
r+2−2s .
≤ C2−jα kvkBp,∞
kukBp,∞
Similarly, we estimate
kI4 kLp ≤ C
X
k(−∆)1−s ukL∞ k∆k vkLp
k≥j
≤ C2−jα
X
2(j−k)α k(−∆)1−s ukL∞ 2kα k∆j vkLp
k≥j
≤ C2
−jα
r+2−2s kvkB α
,
kukBp,∞
p,∞
r+2−2s k∆j vkLp .
kI5 kLp ≤ Ck(−∆)1−s ukL∞ k∆j vkLp ≤ CkukBp,∞
Collecting the estimates above, we obtain
r+1
2jα k[∆j , v](−∆)1−s ukLp ≤ C2j(α+1−2s) k∆j ukLp kvkBp,∞
r+2−2s
r+2−2s k∆j vkLp + CkvkB α
kukBp,∞
+ C2jα kukBp,∞
p,∞
r+1 kuk α+1−2s + CkvkB α
r+2−2s .
≤ CkvkBp,∞
kukBp,∞
Bp,∞
p,∞
This completes the proof.
Proposition 3.4. Let s ∈ [1/2, 1], p = k/l be a rational number with k even, l odd,
α
α
is a solution of (1.1) with u0 ∈ Bp,∞
and α > dp + 1. Assume that u(x, t) ∈ Bp,∞
for t ∈ [0, T ]. Then, when u(x, t) ≥ 0, we can find some C = C(p, α), that for any
t ≤ T,
Z t
α
α
α
kukBp,∞
≤ Cku0 kBp,∞
exp{C
kukBp,∞
dτ }.
(3.6)
0
Proof. Applying ∆j on (1.1), we obtain
X
∂t ∆j u =
[∆j , ∂i (−∆)−s u]∂i u + ∇(−∆)−s u∆j (∇u)
− [∆j , u](−∆)1−s u − u∆j ((−∆)1−s u.
8
X. ZHOU, W. XIAO, J. CHEN
EJDE-2014/199
Multiplying both sides by p∆j u|∆j u|p−2 and integrating over Rd , the equation
becomes
d
k∆j ukpLp
dt
Z
XZ
p∆j u|∆j u|p−2 [∆j , ∂i (−∆)−s u]∂i u + p∆j u|∆j u|p−2 ∇(−∆)−s u∆j (∇u)
=
Z
Z
p−2
1−s
− p∆j u|∆j u| [∆j , u](−∆) u − p∆j u|∆j u|p−2 u∆j ((−∆)1−s u
= J1 + J2 + J3 + J4 .
From Propositions 3.1 and 3.3, we obtain the estimates
p−1
−jα
2
α
α+1−2s ≤ C2
J1 ≤ C2−jα k∆j ukp−1
,
kukBp,∞
k∆j ukL
p kukB α
Lp kukBp,∞
p,∞
−jα
2
α
α+1−2s ≤ C2
k∆j ukp−1
.
kukBp,∞
J3 ≤ C2−jα k∆j ukp−1
α
Lp kukBp,∞
Lp kukBp,∞
It is easy to see that
Z
Z
−s
p
J2 = ∇(−∆) u)∇(|∆j (u)| ) = (−∆)1−s u|∆j u|p dx
p
p−1
α−1 k∆j uk p ≤ Ckuk α+1−2s k∆j uk p k∆j ukLp
≤ Ck(−∆)1−s ukBp,∞
L
L
Bp,∞
p−1
α+1−2s kukB α
k∆j ukL
≤ C2−jα kukBp,∞
p .
p,∞
Using the fact that u ≥ 0 and Propositions 2.5 and 2.6, we estimate
Z
Z
J4 ≤ − p|∆j u|p−2 u(−∆)1−s |∆j u|2 ≤ − u(−∆)1−s |∆j u|p
Z
p
α−1 k∆j uk p
= − (−∆)1−s u|∆j u|p ≤ Ck(−∆)1−s ukBp,∞
L
α+1−2s kukB α
≤ C2−jα kukBp,∞
k∆j ukp−1
Lp .
p,∞
d
Combining the above four estimates, we set dt
. Since u ≥ 0,
kukḂ α ≤ Ckuk2Bp,∞
α
p,∞
it follows
Z
Z
d
d
kukLp =
up dx = p up−1 ut dx
dt
dt
Z
Z
= p up−1 ∇ · (u∇(−∆)−s u) = −(p − 1) ∇up · (∇(−∆)−s u)
Z
= −(p − 1) up ((−∆)1−s u)
α
α+1−2s .
≤ Ckukp−1
kukBp,∞
Lp kukBp,∞
This implies
d
α
α
α
kukBp,∞
≤ CkukBp,∞
kukBp,∞
,
dt
which with the Gronwall’s inequality yield (3.6).
4. Proof of Theorem 1.1
From the definition of Riesz potentials (−∆)−s u = c(n, s)|x|−d+2s ∗u, 0 < 2s < d.
So when d ≥ 2, 21 ≤ s < 1, we define P u = c(n, s)(|x|−d+2s ∗ σ ) ∗ u = (−∆)−s (σ ∗
u) = (−∆)−s u , u = σ ∗ u . Here σ ∈ Cc∞ , is nonnegative, radially symmetric
EJDE-2014/199
FRACTIONAL POROUS MEDIUM EQUATION
9
R
and decreasing, σ = 1, σ = ε−d σ(x/). Then we construct a sequence {u(n) },
defined recursively by solving the following equations
u(1) = S2 (u0 )
∂t u(n+1) = ∇ · (u(n+1) ∇(P u(n) ))
(n+1)
u
(x, 0) =
un+1
0
(4.1)
= Sn+2 u0 .
Since u(n) solves the linear system, we can always find the sequence. Similarly to
the proof of Proposition 3.4, taking ∆j on (4.1), we obtain
X
X
(n+1)
∂t ∆j u(n+1) =
[∆j , ∂i (−∆)−s u(n)
+
∂i (−∆)−s u(n) ∆j (∂i u(n+1) )
]∂i u
− [∆j , u(n+1) ](−∆)1−s u(n)
− u(n+1) ∆j ((−∆)1−s u(n) ).
(4.2)
∆ u(n+1)
Multiplying both sides by |∆jj u(n+1) | , integrating over Rd , we denote each corresponding part in the right side by J10 , J20 , J30 , J40 . Now we obtain the estimates
α
J10 ≤ C2−jα ku(n+1) kB1,∞
ku(n) kB α+1−2s ,
1,∞
α
J30 ≤ C2−jα ku(n+1) kB1,∞
ku(n) kB α+1−2s ,
(4.3)
1,∞
(n)
where we used the fact that ku kB α+1−2s ≤ ku(n) kB α+1−2s . Also, we have
1,∞
1,∞
Z
J20 = ∇(−∆)−s u(n) )∇(|∆j (u(n+1) )|)
Z
(n+1)
= (−∆)1−s u(n)
|dx
|∆j u
(4.4)
α
≤ C2−jα ku(n+1) kB1,∞
ku(n) kB α+1−2s ,
1,∞
and
J40 = −
Z
∆j u(n+1) (n+1)
u
∆j ((−∆)1−s u(n)
)
|∆j u(n+1) |
∆j u(n+1) (n+1)
u
∆j (∆(−∆)−s u(n)
)
|∆j u(n+1) |
Z
∆j u(n+1)
=−
∇u(n+1) ∆j (∇(−∆)−s u(n)
)
|∆j u(n+1) |
Z
=
(4.5)
α
≤ C2−jα ku(n) kB α+1−2s ku(n+1) kB1,∞
.
1,∞
On the other hand,
Z
Z
Z
u(n+1) (n+1)
d
(n+1)
|u
|dx =
u
dx = ∇ · (|u(n+1) |∇(−∆)−s u(n)
) = 0.
dt
|u(n+1) | t
Collecting the estimates above, when s ≥ 1/2, we obtain
d (n+1)
α
α
ku
kB1,∞
≤ Cku(n+1) kB1,∞
ku(n) kB α+1−2s
1,∞
dt
(n+1)
(n)
α
α
≤ Cku
kB1,∞
ku kB1,∞
.
(4.6)
10
X. ZHOU, W. XIAO, J. CHEN
EJDE-2014/199
Now from Gronwall’s inequality we obtain
(n+1)
ku
Z
(n+1)
α
ku0
kB1,∞
t
α
exp{C
ku(n) kB1,∞
)dτ }
0
Z t
α
α
≤ Cku0 kB1,∞
exp{C
ku(n) kB1,∞
)dτ },
α
kB1,∞
≤
0
α
with C independent of n. Defining XT := C([0, T ]; B1,∞
), we have
α
ku(n+1) kXT ≤ Cku0 kB1,∞
exp(CT ku(n) kXT ).
Thus, by the standard induction argument we find
α
α
sup ku(n) kB1,∞
≤ 2Cku0 kB1,∞
,
(4.7)
0≤t≤T0
α
for all n ≥ 1, if exp(2CT0 ku0 kB1,∞
) ≤ 2; namely, T0 ≤
ln 2
2C0 (1+ku0 kB α
1,∞
).
Using
Proposition 3.2, 3.3, and (4.4), (4.5), we see
d (n+1)
α
.
ku
kB α+2s−2 ≤ Cku(n+1) kB α+2s−2 ku(n) kB1,∞
1,∞
1,∞
dt
Thus, from the uniform estimate (4.7), we have
sup k
0≤t≤T0
∂ (n)
u kB α+2s−2 ≤ Cku0 k2B1,∞
.
α
1,∞
∂t
(4.8)
(4.9)
α+2s−2
Let YT := C([0, T ]; B1,∞
). We will prove that the sequence {u(n) } is Cauchy in
YT1 for some T1 ∈ (0, T0 ). Considering the difference u(n+1) − u(n) ,
∂t (u(n+1) − u(n) )
= ∇u(n+1) · ∇(−∆)−s u(n)
− u(n+1) (−∆)1−s u(n) − ∇u(n) · ∇(−∆)−s u(n−1)
+ u(n) (−∆)1−s u(n−1)
= ∇ · ((u(n+1) − u(n) )∇(−∆)−s u(n) ) + ∇ · (u(n) ∇(−∆)−s (u(n)
− u(n−1) )),
with initial datum (u(n+1) − u(n) )(x, 0) = ∆n+1 u0 . Proceeding as in the proof of
(4.6), we obtain the estimate
α
k∇ · (u(n) ∇(−∆)−s (u(n)
− u(n−1)
))kB α+2s−2 ≤ Cku(n) − u(n−1) kB α−1 ku(n) kB1,∞
.
1,∞
1,∞
Proceeding as in the proof of (4.8), we obtain the estimate
(n+1)
α
k∇ · ((u(n+1) − u(n) )∇(−∆)−s u(n)
− u(n) kB α+2s−2 ku(n) kB1,∞
.
)kB α+2s−2 ≤ Cku
1,∞
1,∞
Then we obtain
d (n+1)
α
ku
− u(n) kB α+2s−2 ≤ Cku(n+1) − u(n) kB α+2s−2 ku(n) kB1,∞
1,∞
1,∞
dt
α
+ Cku(n) − u(n−1) kB α+2s−2 ku(n) kB1,∞
,
1,∞
and
k(u(n+1) − u(n) )(x, 0)kB α+2s−2
1,∞
= k∆n+1 u0 kB α+2s−2 = sup 2j(α+2s−2) k∆j ∆n+1 u0 kL1
1,∞
n(2s−2)
≤ C2
sup
n≤j≤n+2
j
jα
2 k∆j u0 kL1
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FRACTIONAL POROUS MEDIUM EQUATION
11
α
≤ C2n(2s−2) ku0 kB1,∞
.
The above inequalities and Gronwall’s inequality imply that
ku(n+1) (t) − u(n) (t)kB α+2s−2
1,∞
α
≤ k(u(n+1) − u(n) )(x, 0)kB α+2s−2 exp(Cku0 kB1,∞
)
1,∞
Z t
α
exp(Cku0 kB1,∞
(t − s))ku(n) − u(n−1) kB α+2s−2 (s)ds
+
0
0 n(2s−2)
≤C2
(4.10)
1,∞
0
(n)
+ C ku
(n−1)
−u
0
kYT1 (exp C T1 − 1),
α
where T1 ∈ [0, T0 ], and the constant C 0 = C 0 (ku0 kB1,∞
). Thus, if C 0 (exp T1 − 1) <
α+2s−2
1
(n)
∞
converges to u ∈ L ([0, T1 ]; B1,∞
) in YT1 . By
2 , we can deduce that u
the well-known interpolation inequality in the Besov spaces we have un → u in
β
L∞ ([0, T1 ]; B1,∞
) for all β ∈ [α + 2s − 2, α]. Moreover, the estimate (4.8) allows us
α+2s−2
to conclude that u ∈ C 1 ([0, T1 ]; B1,∞
). Letting → 0, n → ∞, we find that u is
β
a solution to (1.1) in B1,∞ .
β
Next we prove that the solution is unique. Suppose that u, v in L∞ ([0, T1 ]; B1,∞
)
are two solutions of (1.1) associated with the initial condition u0 , v0 . Then u − v
satisfies the equation
u − v = ∇ · [(u − v)∇(−∆)−s u] + ∇ · [v∇(−∆)−s (u − v)],
(u − v)(x, 0) = u0 − v0 .
Working similarly with (4.10), we can obtain
k(u − v)(t)kB β
1,∞
≤ M 2j(β−α) k(u − v)(x, 0)kB β
1,∞
+ M ku − vkC([0,T1 ];B β
1,∞
00
) (exp C T1 − 1),
for M = M (k(u0 kB β , kv0 kB β ). If M (exp C 00 T1 − 1) < 1, then we get the unique1,∞
1,∞
β
ness of solutions in C([0, T1 ]; B1,∞
). The proof is complete.
5. Proof of Theorem 1.2
When d ≥ 3, we know (−∆)−1 u = c(n, s)|x|−d+2 ∗ u. Let P u = c(n, s)(|x|−d+2 ∗
σ )∗u = (−∆)−1 (σ ∗u) = (−∆)−1 u , where σ is defined in the beginning of Section
4. Consider the successive approximation sequence {u(n) } satisfying
u(1) = S2 (u0 )
(n+1)
∂t u(n+1) = ∇ · (u(n+1) ∇(P u(n)
· ∇(−∆)−1 u(n)
− u(n+1) u(n)
)) = ∇u
(n+1)
u
(x, 0) =
un+1
0
(5.1)
= Sn+2 u0 .
Appling ∆j on (5.1),
∂t ∆j u(n+1) =
X
(n+1)
[∆j , ∂i (−∆)−s u(n)
]∂i u
(n+1)
+ ∇(−∆)−s u(n)
) − ∆j (u(n+1) u(n)
∇∆j (u
).
(5.2)
When p ≥ 2, multiplying both sides of (5.2) by p∆j u(n+1) |∆j u(n+1) |p−2 and integrating over Rd , we obtain
XZ
d
(n+1) p
(n+1)
k∆j u
kLp =
p∆j u(n+1) |∆j u(n+1) |p−2 [∆j , ∂i (−∆)−1 u(n)
]∂i u
dt
12
X. ZHOU, W. XIAO, J. CHEN
Z
+
Z
−
EJDE-2014/199
(n+1)
p∆j u(n+1) |∆j u(n+1) |p−2 ∇(−∆)−1 u(n)
)
∇∆j (u
00
00
00
p∆j u(n+1) |∆j u(n+1) |p−2 ∆j (u(n+1) u(n)
ε ) = J1 + J2 + J3 .
By Hölder’s inequality,
−1 (n)
J100 ≤ Ck∆j u(n+1) kp−1
u ]∂i u(n+1) kLp .
Lp k[∆j , ∂i (−∆)
From the proof of Proposition 2.3 (iv),
(n)
J300 ≤ k∆j u(n+1) kp−1
kL∞ k∆j u(n+1) kLp + ku(n+1) kL∞ k∆j u(n) kLp ].
Lp [ku
Also we obtain
J200 =
Z
Z
=
(n+1) p
∇(−∆)−1 u(n)
)| )
∇(|∆j (u
(n+1) p
u(n)
| dx ≤ ku(n) kL∞ k∆j u(n+1) kpLp .
|∆j u
Now we obtain
d
k∆j u(n+1) kLp ≤ Ck[∆j , ∂i (−∆)−1 u(n) ]∂i u(n+1) kLp
dt
+ Cku(n) kL∞ k∆j u(n+1) kLp + Cku(n+1) kL∞ k∆j u(n) kLp .
When, α > d/p,
d (n+1)
α
α
ku
kḂ α ≤ Cku(n+1) kBp,∞
ku(n) kBp,∞
.
p,∞
dt
Letting p → ∞,
d
k∆j u(n+1) kL∞ ≤ Ck[∆j , ∂i (−∆)−1 u(n) ]∂i u(n+1) kL∞
dt
+ Cku(n) kL∞ k∆j u(n+1) kL∞ + Ck∆j (u(n+1) u(n) )kL∞ .
So for any α > 0,
d (n+1)
α
α
ku
kḂ α ≤ Cku(n+1) kB∞,∞
ku(n) kB∞,∞
.
∞,∞
dt
It is easy to prove that when p = 1,
Z
Z
d
u(n+1) (n+1)
|u(n+1) |dx =
u
dx
dt
|u(n+1) | t
Z
= ∇ · (|u(n+1) |(|x|−d+2 ∗ σ ) ∗ u(n) ) = 0.
Noting u(n+1) (x, 0) = un+1
= Sn+2 u0 , we know ku(n+1) kL1 ≤ C.
0
When p ≥ 2,
Z
Z
d
(n+1)
|u(n+1) |p dx = p |u(n+1) |p−2 u(n+1) ut
dx
dt
Z
= p |u(n+1) |p−2 u(n+1) ∇ · (u(n+1) ∇(−∆)−1 u(n)
)
Z
= −(p − 1) ∇|u(n+1) |p · (∇(−∆)−1 u(n)
)
Z
= (p − 1) |u(n+1) |p u(n)
≤ Cku(n+1) kpLp ku(n) kL∞ .
EJDE-2014/199
FRACTIONAL POROUS MEDIUM EQUATION
13
It means
d (n+1)
α
α
ku
kLp ≤ ku(n+1) kLp ku(n) kL∞ ≤ Cku(n+1) kBp,∞
ku(n) kBp,∞
.
dt
Again letting p → ∞, we obtain
d (n+1)
α
α
ku
kL∞ ≤ Cku(n+1) kB∞,∞
ku(n) kB∞,∞
.
dt
Collecting the estimates above, we now obtain
d (n+1)
d
α
α
α
ku
kBp,∞
≤ Cku(n+1) kBp,∞
ku(n) kBp,∞
, p < ∞, α > ,
dt
p
d (n+1)
α
α
α
ku
kB∞,∞
≤ Cku(n+1) kB∞,∞
ku(n) kB∞,∞
, p = ∞, α > 0.
dt
For any β ∈ (d/p, α), p < ∞, or β ∈ (0, α), p = ∞ we have
d (n+1)
ku
kBp,∞
≤ ku(n+1) kBp,∞
ku(n) kBp,∞
,
β
β
β
dt
and
d (n+1)
ku
− u(n) kBp,∞
≤ ku(n+1) − u(n) kBp,∞
ku(n) kBp,∞
β
β
β
dt
+ ku(n) − u(n−1) kBp,∞
ku(n) kBp,∞
.
β
β
The rest of the proof is similar to the proof of Theorem 1.1, we omit it.
Acknowledgments. This research was supported by the National Natural Science
Foundation of China (11201103).
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Xuhuan Zhou
Department of Mathematics, Zhejiang University, 310027 Hangzhou, China
E-mail address: zhouxuhuan@163.com
Weiliang Xiao
Department of Mathematics, Zhejiang University, 310027 Hangzhou, China
E-mail address: xwltc123@163.com
Jiecheng Chen
Department of Mathematics, Zhejiang Normal University, 321004 Jinhua, China
E-mail address: jcchen@zjnu.edu.cn