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
∂tu− ∇ ·(u∇P u) = 0, P u= (−∆)−su, 0< s≤1, x∈Rd, t >0.
We show that whens∈[1/2,1),α > d+ 1,d≥2, the equation has a unique local in time solution for any initial data inBα1,∞. Moreover, in the critical cases= 1, the solution exists inBp,∞α , 2≤p≤ ∞,α > d/p,d≥3.
1. Introduction Let 0< s≤1, we consider the evolution equation
∂tu− ∇ ·(u∇P u) = 0, P u= (−∆)−su, u(x,0) =u0(x). (1.1) wherex∈Rd,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 V´azquez [2]. They proved the existence of a weak solution when u0 is a bounded function and has exponential decay at infinity. Caffarelli, Soria, and V´azquez [1] studied its regularity theory of the weak solution withu0∈L1∩L∞and the continuity of bounded solutions.
Whens= 1, the equation leads to a mean field equation
ut=∇ ·(u∇P u), P u= (−∆)−1u, u(x,0) =u0(x). (1.2) Equation that was first studied by Lin and Zhang [7]. They proved the existence and uniqueness of positiveL∞ solution in two dimensions. When d≥3, V´azquez and Serfaty [11] studied the existence and uniqueness of the weak solution inL∞ 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 show that the strong solution exists inBαp,∞,α >0.
In this article, we are interested in finding the strong solutions of (1.1) in the Besov spaces Bp,∞α . We will show that when d ≥ 2, s ∈ [1/2,1), α > d+ 1, equation (1.1) has a unique local in time solution for any initial data in Bα1,∞. Whens= 1, the solution extends toBp,∞α ,α > d/p, 2≤p≤ ∞,d≥3. The idea
2000Mathematics 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
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∈ B1,∞α . Then we can findT =T(ku0kB1,∞α ), such that a unique solutionuto(1.1)on [0, T]×Rdexists. And the solution belongs toC1([0, T];B1,∞α+2s−2)∩L∞([0, T];B1,∞β ), andβ∈[α+ 2s−2, α].
Theorem 1.2. Let d≥3, α > d/p when2≤p≤ ∞. Assume that the initial data u0∈Bp,∞α . Then we can find T =T(ku0kBαp,∞), such that a unique solution uto the given mean field equation (1.2) on[0, T]×Rd exists. And the solution belongs toC1([0, T];Bp,∞β )∩L∞([0, T];Bp,∞α ),β ∈(d/p, α),2≤p≤ ∞.
2. Preliminaries
In this section, we recall the definition of the Besov space. We start with a dyadic decomposition ofRd.
Supposeχ∈C0∞(Rd), φ∈C0∞(Rd\ {0}) satisfying suppχ⊂ {ξ∈Rd :|ξ| ≤ 4
3}, suppφ⊂ {ξ∈Rd: 3
4 <|ξ|<8 3}, χ(ξ) +X
j≥0
φ(2−jξ) = 1, ξ∈Rd, X
j∈Z
φ(2−jξ) = 1, ξ∈Rd\{0}.
Define the operators
∆ju=φ(2−jD)u= 2jd Z
h(2jy)u(x−y)dy, Sjf = X
k≤j−1
∆kf =χ(2−jD)u= 2jd Z
g(2jy)u(x−y)dy,
(2.1)
whereg=χ∨andh=φ∨are the inverse Fourier transform ofχandφ, respectively.
It can be easily verified that with our choice ofφ,
∆j∆kf ≡0, if |j−k| ≥2, ∆j(Sk−1f∆kf)≡0, if|j−k| ≥5. (2.2) Definition 2.1. For anyα∈R, andp, q∈[1,∞], the homogeneous Besov spaces B˙p,qr are defined as
B˙αp,q={f ∈ Z0(Rd) :kfkB˙αp,q<∞}.
Here,
kfkB˙αp,q =h X
j∈Z
2jαqk∆jfkqLp
i1/q
, whereq <∞.
Whenq=∞,
kfkB˙αp,∞ = sup
j∈Z
2jαk∆jfkLp.
Z0(Rd) denotes the dual space ofZ(Rd) ={f ∈ S(Rd) :∂γfˆ(0) = 0,∀γ∈Nd} and can be identified by the quotient space ofS0/P with the polynomials spaceP.
Definition 2.2. For anyα∈R, andp, q∈[1,∞], the inhomogeneous Besov space Bp,qr is defined as
Bαp,q={f ∈ S0(Rd) :kfkBαp,q<∞}.
Here,
kfkBαp,q=X∞
j≥0
2jαqk∆jfkqLp
1/q
+kS0(f)kLp, whenq <∞.
Whenq=∞,
kfkBp,∞α = sup
j≥0
2jαk∆jfkLp+kS0(f)kLp. Let us state some basic properties of the Besov spaces.
Proposition 2.3. Let s∈R,1≤p≤ ∞,1≤q≤ ∞.
(i) If α >0, thenBp,qα = ˙Bαp,q∩Lp, andkfkBαp,q=kfkB˙p,qα +kfkLp;
(ii) If α1 ≤α2, thenBαp,q2 ⊂Bp,qα1. If 1≤q1≤q2 ≤ ∞, thenB˙p,qα 1 ⊂B˙p,qα 2 and Bp,qα 1⊂Bp,qα 2;
(iii) If α > dp, thenBp,qα ,→L∞. If p1≤p2, α1−pd
1 > α2−pd
2, thenBpα1
1,q1 ,→ Bpα2
2,q2,Bp,min(p,2)α ,→Hpα,→Bαp,max(p,2); (iv) If α >0,p≥1, thenkuvkBα
p,∞ ≤CkukL∞kvkBα
p,∞+CkukBα
p,∞kvkL∞. We now turn to Bernstein’s inequalities. When the Fourier transform of a func- tion is supported on a ball or an annulus, the Lp-norms of the derivatives of the function can be bounded in terms of theLp-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 andsupp ˆf ⊂ {ξ∈ Rd:|ξ| ≤K2j} for someK >0 and integerj, then
k(−∆)αDγfkLq ≤C2j(2α+|γ|)+jd(1p−1q)
kfkLp.
(2) If α∈Rand supp ˆf ⊂ {ξ∈Rd :K12j ≤ |ξ| ≤K22j} for some K1, K2>0 and integerj, then
C2j(2α+|γ|)+jd(p1−1q)
kfkLp≤ k(−∆)αDγfkLq ≤C2e j(2α+|γ|)+jd(1p−q1)
kfkLp, whereC andCe are positive constants independent of j.
Next we state two pointwise inequalities which were proved in [5, 13].
Proposition 2.5. Let 0 ≤α≤ 1, f ∈C2(Rd) decay sufficiently fast at infinity.
Then for anyx∈Rd,
2f(x)(−∆)αf(x)≥(−∆)αf2(x).
Proposition 2.6. Let 0 ≤α ≤ 1, p1 = kl1
1 ≥0, p2 = kl2
2 ≥ 1 be rational num- bers with l1, l2 odd, and k1l1+k2l2 even. Then for any f ∈ C2(Rd) that decays sufficiently fast at infinity, and for anyx∈Rd,
(p1+p2)fp1(x)(−∆)αfp2(x)≥p2(−∆)αfp1+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 constantC such that
2jαk[∆j, ∂i(−∆)−su]∂ivkLp ≤CkvkBr+1p,∞kukBp,∞α+1−2s+CkukBp,∞r+2−2skvkBp,∞α , (3.1) where the brackets[,]represents the commutator, namely
[∆j, ∂i(−∆)−su]∂iv= ∆j(∂i(−∆)−su)∂iv)−∂i(−∆)−su∆j(∂iv).
Proof. Using Bony’s para-product decomposition, we have [∆j, ∂i(−∆)−su]∂iv=L1+L2+L3+L4+L5, where
L1= X
|k−j|≤4
∆j[Sk−1(∂i(−∆)−su)∆k(∂iv)]−Sk−1(∂i(−∆)−su)∆k(∆j(∂iv)), L2= X
|k−j|≤4
∆j[Sk−1(∂iv)∆k(∂i(−∆)−su)], L3= X
k≥j−2
∆j(∆k(∂i(−∆)−su)∆ek(∂iv)), L4=X
k
Sk−1(∆j(∂iv))∆k(∂i(−∆)−su), L5= X
|j0−j00|≤1
∆j0(∆j(∂iv))∆j00(∂i(−∆)−su).
We shall estimate the above terms separately. First observe L1= X
|k−j|≤4
2jd Z
h(2j(x−y))
Sk−1(∂i(−∆)−su)(y)
−Sk−1(∂i(−∆)−su)(x)
∆k(∆j(∂iv))(y)dy.
By Young’s inequality and Bernstein’s inequality, kL1kLp≤C X
|k−j|≤4
2−jk∇∂i(−∆)−sukL∞k∆j(∂iv)kLp
Z
|y||h(y)|dy
≤C2−j2jk(−∆)1−sukL∞k∆jvkLp
≤Ck∆jvkLpkukBr+2−2s
p,∞ .
Similarly,
kL2kLp≤C2j(1−2s)k∇vkL∞k∆jukLp≤C2−jαkvkBr+1
p,∞kukBα+1−2s
p,∞ .
We can also estimate, kL3kLp ≤C X
k≥j−2
k∆k(∂i(−∆)−su)kLpk∇vkL∞
≤C2−jα X
k≥j−2
2(j−k)α2k(α+1−2s)k∆kukLpkvkBr+1p,∞
≤C2−jαkukBα+1−2s
p,∞ kvkBr+1 p,∞.
To estimateL4, by the definition ofSjand ∆j, we can observe that onlyksatisfying k≥j survive. Thus
kL4kLp≤X
k≥j
Ck∇vkL∞2k(1−2s)k∆kukLp
≤C2−jαX
k≥j
2(j−k)αkvkBr+1
p,∞kukBα+1−2s
p,∞
≤C2−jαkvkBr+1
p,∞kukBα+1−2s
p,∞ .
Since ∆k∆j = 0, for|j−k| ≥2, we have
kL5kLp≤Ck∇vkL∞2j(1−2s)k∆jukLp≤C2−jαkvkBr+1
p,∞kukBα+1−2s
p,∞ .
Collecting the estimates above, we obtain 2jαk[∆j, ∂i(−∆)−su]∂ivkLp
≤C2jαk∆jvkLpkukBr+2−2s
p,∞ + 2j(α+1−2s)kvkBr+1
p,∞k∆jukLp+kukBα+1−2s
p,∞ kvkBr+1 p,∞
≤ kvkBα
p,∞kukBr+2−2s
p,∞ +kukBα+1−2s
p,∞ kvkBr+1 p,∞.
This completes the proof.
Proposition 3.2. Let α, s, p, r be as in Proposition 3.1. Then there exists a con- stantC such that
2jαk[∆j, ∂i(−∆)−su]∂ivkLp ≤CkvkBrp,∞kukBp,∞α+2−2s+CkukBp,∞r+2−2skvkBp,∞α . (3.2) Whenp=∞, this inequality becomes: for anyr >0,
2jαk[∆j, ∂i(−∆)−su]∂ivkL∞≤CkvkBr∞,∞kukBα+2−2s
∞,∞ +CkukBr+2−2s
∞,∞ kvkBα∞,∞. (3.3) Proof. We want to give a new estimate of the commutator in Proposition 3.1.
Following the above proof, the estimate ofL1unchanged, we give different bounds forL2, L3, L4, L5. First,
L2= X
|k−j|≤4
2jd Z
h(2j(x−y))(Sk−1∂iv)(y)∆k(∂i(−∆)−su)(y)dy
= X
|k−j|≤4
2jd Z
∂ih(2j(x−y))2j(Sk−1v)(y)∆k(∂i(−∆)−su)(y)dy
− X
|k−j|≤4
2jd Z
h(2j(x−y))(Sk−1v)(y)∆k(∂ii(−∆)−su)(y)dy.
So we obtain
kL2kLp ≤C22−2s(k∂ihkL1+khkL1)kvkL∞k∆jukLp
≤C2−jαkukBα+2−2s
p,∞ kvkBrp,∞.
Similarly,
kL3kLp≤C X
k≥j−2
2jd Z
∂ih(2j(x−y))2j∆k(∂i(−∆)−su)(y)∆kv(y)dy
−2jd Z
h(2j(x−y))∆k(∂ii(−∆)−su)(y)∆kv(y)dy.
Hence we obtain, kL3kLp ≤C X
k≥j−2
(2(j−k)(α+1)+ 2(j−k)α)2k(α+2−2s)kvkL∞k∆kukLp
≤C2−jαkukBα+2−2s
p,∞ kvkBp,∞r . Also,
kL4kLp≤X
k≥j
C2jk∆jvkL∞2k(1−2s)k∆kukLp
≤C2−jαX
k≥j
2(j−k)(α+1)2k(α+2−2s)kvkBp,∞r k∆kukLp
≤C2−jαkvkBp,∞r kukBα+2−2s
p,∞ .
Finally,
kL5kLp≤Ck∆jvkL∞2j(2−2s)k∆jukLp≤C2−jαkvkBr
p,∞kukBα+2−2s
p,∞ .
Collecting the estimates above, we can obtain
2jαk[∆j, ∂i(−∆)−su]∂ivkLp≤C2jαk∆jvkLpkukBr+2−2s
p,∞ +kvkBp,∞r kukBα+2−2s
p,∞
≤ kvkBp,∞α kukBr+2−2s
p,∞ +kukBα+2−2s
p,∞ kvkBp,∞r .
This completes the proof.
Proposition 3.3. Let α >0,s ∈(0,1),p∈ [1,∞]. Assume r > dp. Then there exists a constantC such that
2jαk[∆j, v](−∆)1−sukLp ≤CkvkBr+1
p,∞kukBα+1−2s
p,∞ +CkukBr+2−2s
p,∞ kvkBp,∞α , (3.4) where the brackets[,]represents the commutator,
[∆j, v](−∆)1−su= ∆j(v(−∆)1−su)−v∆j((−∆)1−su). (3.5) Proof. This proposition is proved similarly to Proposition 3.1. We start the proof by writing
[∆j, v](−∆)1−su=I1+I2+I3+I4+I5, where
I1= X
|k−j|≤4
∆j(Sk−1v∆k(−∆)1−su)−Sk−1v∆k(∆j(−∆)1−su), I2= X
|k−j|≤4
∆j(Sk−1(−∆)1−su)∆kv), I3= X
k≥j−2
∆j(∆kv)e∆k(−∆)1−su), I4=X
k
Sk−1(∆j(−∆)1−su)∆kv,
I5= X
|j0−j00|≤1
∆j0v∆j00(∆k(−∆)1−su).
Similar to the proof for Proposition 3.1, first we observe that kI1kLp =k X
|k−j|≤4
2jd Z
h(2j(x−y))(Sk−1v(y)−Sk−1v(x))∆k(−∆)1−su)(y)dykLp
≤C X
|k−j|≤4
2−jk∇vkL∞k∆j(−∆)1−sukLp
Z
|y||h(y)|dy
≤C2−j22j(1−s)k∇vkL∞k∆jukLp
≤C2j(1−2s)k∆jukLpkvkBr+1 p,∞. Also we obtain
kI2kLp≤Ck(−∆)1−sukL∞k∆jvkLp≤CkukBr+2−2s
p,∞ k∆jvkLp, kI3kLp≤C X
k≥j−2
k∆kvkLp2k(2−2s)k∆kukL∞
≤C2−jα X
k≥j−2
2(j−k)α2kαk∆kvkLp2k(2−2s)k∆kukL∞
≤C2−jαkvkBα
p,∞kukBr+2−2s
p,∞ .
Similarly, we estimate kI4kLp≤CX
k≥j
k(−∆)1−sukL∞k∆kvkLp
≤C2−jαX
k≥j
2(j−k)αk(−∆)1−sukL∞2kαk∆jvkLp
≤C2−jαkukBr+2−2sp,∞ kvkBαp,∞,
kI5kLp≤Ck(−∆)1−sukL∞k∆jvkLp≤CkukBr+2−2s
p,∞ k∆jvkLp. Collecting the estimates above, we obtain
2jαk[∆j, v](−∆)1−sukLp≤C2j(α+1−2s)k∆jukLpkvkBr+1 p,∞
+C2jαkukBr+2−2s
p,∞ k∆jvkLp+CkvkBα
p,∞kukBr+2−2s p,∞
≤CkvkBr+1
p,∞kukBα+1−2s
p,∞ +CkvkBα
p,∞kukBr+2−2s
p,∞ .
This completes the proof.
Proposition 3.4. Lets∈[1/2,1],p=k/lbe a rational number withkeven,lodd, andα > dp + 1. Assume that u(x, t)∈Bαp,∞ is a solution of (1.1)with u0 ∈Bαp,∞
fort∈[0, T]. Then, when u(x, t)≥0, we can find someC=C(p, α), that for any t≤T,
kukBαp,∞ ≤Cku0kBp,∞α exp{C Z t
0
kukBp,∞α dτ}. (3.6) Proof. Applying ∆j on (1.1), we obtain
∂t∆ju=X
[∆j, ∂i(−∆)−su]∂iu+∇(−∆)−su∆j(∇u)
−[∆j, u](−∆)1−su−u∆j((−∆)1−su.
Multiplying both sides by p∆ju|∆ju|p−2 and integrating over Rd, the equation becomes
d
dtk∆jukpLp
=XZ
p∆ju|∆ju|p−2[∆j, ∂i(−∆)−su]∂iu+ Z
p∆ju|∆ju|p−2∇(−∆)−su∆j(∇u)
− Z
p∆ju|∆ju|p−2[∆j, u](−∆)1−su− Z
p∆ju|∆ju|p−2u∆j((−∆)1−su
=J1+J2+J3+J4.
From Propositions 3.1 and 3.3, we obtain the estimates
J1≤C2−jαk∆jukp−1Lp kukBαp,∞kukBα+1−2sp,∞ ≤C2−jαk∆jukp−1Lp kuk2Bα p,∞, J3≤C2−jαk∆jukp−1Lp kukBαp,∞kukBα+1−2sp,∞ ≤C2−jαk∆jukp−1Lp kuk2Bα
p,∞. It is easy to see that
J2= Z
∇(−∆)−su)∇(|∆j(u)|p) = Z
(−∆)1−su|∆ju|pdx
≤Ck(−∆)1−sukBα−1
p,∞k∆jukpLp≤CkukBα+1−2s
p,∞ k∆jukp−1Lp k∆jukLp
≤C2−jαkukBα+1−2s p,∞ kukBα
p,∞k∆jukp−1Lp .
Using the fact thatu≥0 and Propositions 2.5 and 2.6, we estimate J4≤ −
Z
p|∆ju|p−2u(−∆)1−s|∆ju|2≤ − Z
u(−∆)1−s|∆ju|p
=− Z
(−∆)1−su|∆ju|p≤Ck(−∆)1−sukBα−1
p,∞k∆jukpLp
≤C2−jαkukBp,∞α+1−2skukBp,∞α k∆jukp−1Lp .
Combining the above four estimates, we set dtdkukB˙p,∞α ≤Ckuk2Bα
p,∞. Sinceu≥0, it follows
d
dtkukLp= d dt
Z
updx=p Z
up−1utdx
=p Z
up−1∇ ·(u∇(−∆)−su) =−(p−1) Z
∇up·(∇(−∆)−su)
=−(p−1) Z
up((−∆)1−su)
≤Ckukp−1Lp kukBp,∞α kukBα+1−2s
p,∞ .
This implies
d
dtkukBp,∞α ≤CkukBp,∞α kukBp,∞α ,
which with the Gronwall’s inequality yield (3.6).
4. Proof of Theorem 1.1
From the definition of Riesz potentials (−∆)−su=c(n, s)|x|−d+2s∗u, 0<2s < d.
So whend≥2, 12≤s <1, we definePu=c(n, s)(|x|−d+2s∗σ)∗u= (−∆)−s(σ∗ u) = (−∆)−su, u =σ∗u. Here σ ∈ Cc∞, is nonnegative, radially symmetric
and decreasing, R
σ = 1, σ =ε−dσ(x/). Then we construct a sequence {u(n)}, defined recursively by solving the following equations
u(1)=S2(u0)
∂tu(n+1)=∇ ·(u(n+1)∇(Pu(n))) u(n+1)(x,0) =un+10 =Sn+2u0.
(4.1)
Sinceu(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
∂t∆ju(n+1)=X
[∆j, ∂i(−∆)−su(n) ]∂iu(n+1)+X
∂i(−∆)−su(n) ∆j(∂iu(n+1))
−[∆j, u(n+1)](−∆)1−su(n) −u(n+1)∆j((−∆)1−su(n) ).
(4.2) Multiplying both sides by ∆ju(n+1)
|∆ju(n+1)|, integrating over Rd, we denote each corre- sponding part in the right side byJ10, J20, J30, J40. Now we obtain the estimates
J10 ≤C2−jαku(n+1)kBα
1,∞ku(n)kBα+1−2s 1,∞ , J30 ≤C2−jαku(n+1)kBα1,∞ku(n)kBα+1−2s
1,∞ ,
(4.3)
where we used the fact thatku(n) kBα+1−2s
1,∞ ≤ ku(n)kBα+1−2s
1,∞ . Also, we have J20 =
Z
∇(−∆)−su(n) )∇(|∆j(u(n+1))|)
= Z
(−∆)1−su(n) |∆ju(n+1)|dx
≤C2−jαku(n+1)kB1,∞α ku(n)kBα+1−2s 1,∞ ,
(4.4)
and
J40 =−
Z ∆ju(n+1)
|∆ju(n+1)|u(n+1)∆j((−∆)1−su(n) )
=
Z ∆ju(n+1)
|∆ju(n+1)|u(n+1)∆j(∆(−∆)−su(n) )
=−
Z ∆ju(n+1)
|∆ju(n+1)|∇u(n+1)∆j(∇(−∆)−su(n) )
≤C2−jαku(n)kBα+1−2s
1,∞ ku(n+1)kB1,∞α .
(4.5)
On the other hand, d
dt Z
|u(n+1)|dx=
Z u(n+1)
|u(n+1)|u(n+1)t dx= Z
∇ ·(|u(n+1)|∇(−∆)−su(n) ) = 0.
Collecting the estimates above, whens≥1/2, we obtain d
dtku(n+1)kB1,∞α ≤Cku(n+1)kB1,∞α ku(n)kBα+1−2s 1,∞
≤Cku(n+1)kB1,∞α ku(n)kBα1,∞.
(4.6)
Now from Gronwall’s inequality we obtain ku(n+1)kBα
1,∞ ≤ ku(n+1)0 kBα
1,∞exp{C Z t
0
ku(n)kBα
1,∞)dτ}
≤Cku0kB1,∞α exp{C Z t
0
ku(n)kB1,∞α )dτ}, withC independent ofn. Defining XT :=C([0, T];B1,∞α ), we have
ku(n+1)kXT ≤Cku0kB1,∞α exp(CTku(n)kXT).
Thus, by the standard induction argument we find sup
0≤t≤T0
ku(n)kB1,∞α ≤2Cku0kBα1,∞, (4.7) for all n ≥ 1, if exp(2CT0ku0kBα1,∞) ≤ 2; namely, T0 ≤ 2C ln 2
0(1+ku0kBα
1,∞). Using Proposition 3.2, 3.3, and (4.4), (4.5), we see
d
dtku(n+1)kBα+2s−2
1,∞ ≤Cku(n+1)kBα+2s−2
1,∞ ku(n)kBα
1,∞. (4.8)
Thus, from the uniform estimate (4.7), we have sup
0≤t≤T0
k∂
∂tu(n)kBα+2s−2
1,∞ ≤Cku0k2Bα
1,∞. (4.9)
LetYT :=C([0, T];B1,∞α+2s−2). We will prove that the sequence{u(n)} is Cauchy in YT1 for some T1∈(0, T0). Considering the differenceu(n+1)−u(n),
∂t(u(n+1)−u(n))
=∇u(n+1)· ∇(−∆)−su(n) −u(n+1)(−∆)1−su(n) − ∇u(n)· ∇(−∆)−su(n−1) +u(n)(−∆)1−su(n−1)
=∇ ·((u(n+1)−u(n))∇(−∆)−su(n) ) +∇ ·(u(n)∇(−∆)−s(u(n) −u(n−1) )), with initial datum (u(n+1)−u(n))(x,0) = ∆n+1u0. Proceeding as in the proof of (4.6), we obtain the estimate
k∇ ·(u(n)∇(−∆)−s(u(n) −u(n−1) ))kBα+2s−2
1,∞ ≤Cku(n)−u(n−1)kBα−1
1,∞ku(n)kB1,∞α . Proceeding as in the proof of (4.8), we obtain the estimate
k∇ ·((u(n+1)−u(n))∇(−∆)−su(n) )kBα+2s−2
1,∞ ≤Cku(n+1)−u(n)kBα+2s−2
1,∞ ku(n)kB1,∞α . Then we obtain
d
dtku(n+1)−u(n)kBα+2s−2
1,∞ ≤Cku(n+1)−u(n)kBα+2s−2
1,∞ ku(n)kBα1,∞
+Cku(n)−u(n−1)kBα+2s−2
1,∞ ku(n)kBα1,∞, and
k(u(n+1)−u(n))(x,0)kBα+2s−2 1,∞
=k∆n+1u0kBα+2s−2
1,∞ = sup
j
2j(α+2s−2)k∆j∆n+1u0kL1
≤C2n(2s−2) sup
n≤j≤n+2
2jαk∆ju0kL1
≤C2n(2s−2)ku0kBα1,∞.
The above inequalities and Gronwall’s inequality imply that ku(n+1)(t)−u(n)(t)kBα+2s−2
1,∞
≤ k(u(n+1)−u(n))(x,0)kBα+2s−2
1,∞ exp(Cku0kBα1,∞) +
Z t
0
exp(Cku0kB1,∞α (t−s))ku(n)−u(n−1)kBα+2s−2 1,∞ (s)ds
≤C02n(2s−2)+C0ku(n)−u(n−1)kYT1(expC0T1−1),
(4.10)
whereT1∈[0, T0], and the constantC0=C0(ku0kBα1,∞). Thus, ifC0(expT1−1)<
1
2, we can deduce that u(n) converges to u ∈ L∞([0, T1];B1,∞α+2s−2) in YT1. By the well-known interpolation inequality in the Besov spaces we have un → u in L∞([0, T1];B1,∞β ) for allβ ∈[α+ 2s−2, α]. Moreover, the estimate (4.8) allows us to conclude thatu∈C1([0, T1];B1,∞α+2s−2). Letting→0, n→ ∞,we find thatuis a solution to (1.1) inBβ1,∞.
Next we prove that the solution is unique. Suppose thatu,vinL∞([0, T1];B1,∞β ) are two solutions of (1.1) associated with the initial condition u0, v0. Thenu−v satisfies the equation
u−v=∇ ·[(u−v)∇(−∆)−su] +∇ ·[v∇(−∆)−s(u−v)], (u−v)(x,0) =u0−v0.
Working similarly with (4.10), we can obtain k(u−v)(t)kBβ
1,∞≤M2j(β−α)k(u−v)(x,0)kBβ 1,∞
+Mku−vkC([0,T
1];Bβ1,∞)(expC00T1−1), forM =M(k(u0kBβ
1,∞,kv0kBβ
1,∞). IfM(expC00T1−1)<1, then we get the unique- ness of solutions inC([0, T1];B1,∞β ). The proof is complete.
5. Proof of Theorem 1.2
Whend≥3, we know (−∆)−1u=c(n, s)|x|−d+2∗u. LetPu=c(n, s)(|x|−d+2∗ σ)∗u= (−∆)−1(σ∗u) = (−∆)−1u, whereσis defined in the beginning of Section 4. Consider the successive approximation sequence{u(n)}satisfying
u(1)=S2(u0)
∂tu(n+1)=∇ ·(u(n+1)∇(Pu(n) )) =∇u(n+1)· ∇(−∆)−1u(n) −u(n+1)u(n) u(n+1)(x,0) =un+10 =Sn+2u0.
(5.1)
Appling ∆j on (5.1),
∂t∆ju(n+1)=X
[∆j, ∂i(−∆)−su(n) ]∂iu(n+1)
+∇(−∆)−su(n) ∇∆j(u(n+1))−∆j(u(n+1)u(n) ).
(5.2) Whenp≥2, multiplying both sides of (5.2) byp∆ju(n+1)|∆ju(n+1)|p−2 and inte- grating overRd, we obtain
d
dtk∆ju(n+1)kpLp =XZ
p∆ju(n+1)|∆ju(n+1)|p−2[∆j, ∂i(−∆)−1u(n) ]∂iu(n+1)
+ Z
p∆ju(n+1)|∆ju(n+1)|p−2∇(−∆)−1u(n) ∇∆j(u(n+1))
− Z
p∆ju(n+1)|∆ju(n+1)|p−2∆j(u(n+1)u(n)ε ) =J100+J200+J300. By H¨older’s inequality,
J100≤Ck∆ju(n+1)kp−1Lp k[∆j, ∂i(−∆)−1u(n) ]∂iu(n+1)kLp. From the proof of Proposition 2.3 (iv),
J300≤ k∆ju(n+1)kp−1Lp [ku(n)kL∞k∆ju(n+1)kLp+ku(n+1)kL∞k∆ju(n)kLp].
Also we obtain J200=
Z
∇(−∆)−1u(n) ∇(|∆j(u(n+1))|p)
= Z
u(n) |∆ju(n+1)|pdx≤ ku(n)kL∞k∆ju(n+1)kpLp. Now we obtain
d
dtk∆ju(n+1)kLp≤Ck[∆j, ∂i(−∆)−1u(n)]∂iu(n+1)kLp
+Cku(n)kL∞k∆ju(n+1)kLp+Cku(n+1)kL∞k∆ju(n)kLp. When,α > d/p,
d
dtku(n+1)kB˙αp,∞ ≤Cku(n+1)kBα
p,∞ku(n)kBα p,∞. Lettingp→ ∞,
d
dtk∆ju(n+1)kL∞≤Ck[∆j, ∂i(−∆)−1u(n)]∂iu(n+1)kL∞
+Cku(n)kL∞k∆ju(n+1)kL∞+Ck∆j(u(n+1)u(n))kL∞. So for anyα >0,
d
dtku(n+1)kB˙α∞,∞ ≤Cku(n+1)kBα
∞,∞ku(n)kBα
∞,∞. It is easy to prove that whenp= 1,
d dt
Z
|u(n+1)|dx=
Z u(n+1)
|u(n+1)|u(n+1)t dx
= Z
∇ ·(|u(n+1)|(|x|−d+2∗σ)∗u(n)) = 0.
Notingu(n+1)(x,0) =un+10 =Sn+2u0, we knowku(n+1)kL1≤C.
Whenp≥2, d dt
Z
|u(n+1)|pdx=p Z
|u(n+1)|p−2u(n+1)u(n+1)t dx
=p Z
|u(n+1)|p−2u(n+1)∇ ·(u(n+1)∇(−∆)−1u(n) )
=−(p−1) Z
∇|u(n+1)|p·(∇(−∆)−1u(n) )
= (p−1) Z
|u(n+1)|pu(n) ≤Cku(n+1)kpLpku(n)kL∞.
It means d
dtku(n+1)kLp≤ ku(n+1)kLpku(n)kL∞ ≤Cku(n+1)kBαp,∞ku(n)kBp,∞α . Again lettingp→ ∞, we obtain
d
dtku(n+1)kL∞ ≤Cku(n+1)kBα
∞,∞ku(n)kBα
∞,∞. Collecting the estimates above, we now obtain
d
dtku(n+1)kBαp,∞≤Cku(n+1)kBp,∞α ku(n)kBαp,∞, p <∞, α > d p, d
dtku(n+1)kBα∞,∞ ≤Cku(n+1)kBα∞,∞ku(n)kB∞,∞α , p=∞, α >0.
For anyβ∈(d/p, α),p <∞, orβ ∈(0, α),p=∞we have d
dtku(n+1)kBβ
p,∞ ≤ ku(n+1)kBβ
p,∞ku(n)kBβ p,∞, and
d
dtku(n+1)−u(n)kBβ
p,∞≤ ku(n+1)−u(n)kBβ
p,∞ku(n)kBβ p,∞
+ku(n)−u(n−1)kBβ
p,∞ku(n)kBβ p,∞. 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:[email protected]
Weiliang Xiao
Department of Mathematics, Zhejiang University, 310027 Hangzhou, China E-mail address:[email protected]
Jiecheng Chen
Department of Mathematics, Zhejiang Normal University, 321004 Jinhua, China E-mail address:[email protected]