ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu
REGULARITY CRITERIA FOR THE WAVE MAP AND RELATED SYSTEMS
JISHAN FAN, YONG ZHOU
Abstract. We obtain some regularity criteria for the wave map, a liquid crystals model, and the Hall-MHD with ion-slip effect.
1. Introduction
First, we consider thenD wave maps d :R1+n → Sm ⊂R1+m which obey the nonlinear wave equation
∂2td−∆d=d(|∇d|2− |∂td|2) (1.1) with the initial conditions
(d, ∂td)(·,0) = (d0, d1), d0∈Sm, d0·d1= 0. (1.2) Wave maps have wide applications in physics from the harmonic gauge in general relativity to the nonlinearσ-models in particle physics.
Local well-posedness of (1.1) (1.2) has been proved by Tao [20]. Shatah [19]
showed that solutions to the Cauchy problem for wave maps may blow up in finite time. However, some smallness assumption on the initial data or integrability condition on the solution itself are sufficient to guarantee the regularity. Fan and Ozawa [11] obtained the regularity criterion
∇d, ∂td∈L1(0, T; ˙B0∞,∞(Rn)) (1.3) whenn= 2.
The first aim of this article is to prove a following regularity criterion when n≥3.
Theorem 1.1. Letn≥3and(∇d0, d1)∈H1+s(Rn)withs > n2,|d0|= 1,d0·d1= 0 anddbe a smooth solution of (1.1),(1.2). If (1.3)and∂td∈L∞(0, T;Ln(Rn)) hold true with0< T <∞, then the solution dcan be extended beyondT >0.
Next, we consider the liquid crystals model [2, 3, 4, 16]:
∂tu+u· ∇u+∇π−∆u=−∇ ·(∇d ∇d), (1.4)
∂td+u· ∇d−∆d=d|∇d|2, |d|= 1, (1.5)
divu= 0, (1.6)
2010Mathematics Subject Classification. 35K55, 35Q35, 70S15.
Key words and phrases. Regularity criterion; wave map; liquid crystals; Hall-MHD.
c
2016 Texas State University.
Submitted February 18, 2016. Published March 29, 2016.
1
(u, d)(·,0) = (u0, d0) inRn, |d0|= 1. (1.7) Hereuis the velocity,πis the pressure,dis the direction vector, and (∇d∇d)i,j:=
P
k∂idk∂jdk, and hence
∇ ·(∇d ∇d) =X
k
∆dk∇dk+1
2∇|∇d|2. Ifu= 0, then (1.5) is the harmonic heat flow.
Fan-Gao-Guo [9] proved the blow-up criterion
u,∇d∈L2(0, T; ˙B0∞,∞) (1.8) when n = 3. One can find other related results in [8, 24] and references therein.
We will prove the following theorem.
Theorem 1.2. Letn≥3ands > n2 be an integer. Letu0 andd0satisfyu0,∇d0∈ Hs,divu0 = 0, and |d0| = 1 in Rn. Let (u, d) be a local strong solution to the problem (1.4)-(1.7). If∇uand∇2dsatisfy
∇u,∇2d∈L2−α2 (0, T; ˙B∞,∞−α (Rn)) (1.9) with 0 < α <1 and 0 < T <∞, then the solution (u, d) can be extended beyond T >0.
Also we consider the incompressible MHD with the Hall or ion-slip system
∂tu+u· ∇u+∇ π+1
2|b|2
−∆u=b· ∇b, (1.10)
∂tb+u· ∇b−b· ∇u+hrot(rotb×b)−γrot[(rotb×b)×b] = ∆b, (1.11)
divu= divb= 0, (1.12)
(u, b)(·,0) = (u0, b0) inR3. (1.13) Here bis the magnetic field. his the Hall effect coefficient, andγ ≥0 the ion-slip effect coefficient, respectively.
Applications of the Hall-MHD system cover a very wide range of physical sub- jects, such as, magnetic reconnection in space plasmas, star formation, neutron stars, and geo-dynamos.
Very recently, Zhang [23] obtained the regularity criterion
u∈L1−α2 (0, T; ˙B∞,∞−α ), ∇b∈L1−β2 (0, T; ˙B−β∞,∞) (1.14) with−1< α <1 and 0< β <1 whenh= 1 andγ= 0.
Local well-posedness of strong solutions to (1.10)-(1.13) has been proved by Fan, Jia, Nakamura and Zhou [10], they also obtained the regularity criterion
u∈Lq−32q (0, T;Lq), b∈L∞(0, T;L∞), ∇b∈Lp−32p (0, T;Lp) (1.15) with 3< p, q≤ ∞. For standard Hall-MHD system we refer to [1, 5, 6, 7, 13, 21, 22]
and references therein.
By the method in [23], we will refine (1.15) as follows.
Theorem 1.3. Let u0, b0 ∈ H2 with divu0 = divb0 = 0 in R3. Let (u, b) be a local strong solution to the problem (1.10)-(1.13). If u and b satisfy (1.14) and b∈L∞(0, T;L∞)with0< T <∞, then the solution (u, b)can be extended beyond T >0.
In the following proofs, we use the logarithmic Sobolev inequality [15]:
k∇dkL∞ ≤C(1 +k∇dkB˙0∞,∞log(e+k∇dkH1+s)), (1.16) k∂tdkL∞ ≤C(1 +k∂tdkB˙∞,∞0 log(e+k∂tdkH1+s)) (1.17) for s > n2 −1, and the bilinear product and commutator estimates due to Kato- Ponce [14]:
kΛs(f g)kLp≤C(kΛsfkLp1kgkLq1 +kfkLp2kΛsgkLq2), (1.18) kΛs(f g)−fΛsgkLp≤C(k∇fkLp1kΛs−1gkLq1 +kΛsfkLp2kgkLq2), (1.19) withs >0, Λ := (−∆)12 and 1p = p1
1 +q1
1 =p1
2 +q1
2. We also use the Gagliardo-Nirenberg inequalities
k∇dk2L2p≤CkdkL∞k∇2dkLp, (1.20) k∇2dkLp≤Ck∇dk1−θL∞kΛ2+sdkθL2, (1.21) kΛ1+sdk
L
2p
p−2 ≤Ck∇dkθL∞kΛ2+sdk1−θL2 (1.22) withp:= 2s+ 2 and θ:= 1/(1 +s).
We also use the improved Gagliardo-Nirenberg inequalities [12, 17, 18]:
k∇ukLq1 ≤Ck∇uk1−θ˙ 1
B∞,∞−α kukθ˙1
Hs+α, (1.23)
kΛsuk
L
2q1
q1−2 ≤Ck∇ukθ˙1
B−α∞,∞kuk1−θ˙ 1
Hs+α, (1.24)
withq1:=2(s−1+2α)α andθ1:= 2/q1, and k∇dkLq2 ≤Ck∇dk1−θ˙ 2
B−α∞,∞k∇dkθ˙2
Hs+α, (1.25)
kΛs∇dk
L
2q2
q2−2 ≤Ck∇dkθ˙2
B∞,∞−α k∇dk1−θ˙ 2
Hs+α, (1.26)
withq2:=2(s+2α)α andθ2:= q2
2,
k∇dkB˙−α∞,∞≤Ckdk
1 2−α
L∞ k∇2dk
1−α 2−α
B˙−α∞,∞, (1.27) and
kDkukLpk ≤Ck∇uk1−˙ θ˜k
B−α∞,∞kuk1−˙ θ˜k
Hs+α, (1.28)
kDs+2−kdk
L
2pk pk−2
≤Ck∇2dkθ˜˙k
B−α∞,∞
k∇dk1−˙ θ˜k
Hs+α, (1.29)
withpk:= ˜2
θk and ˜θk:= s+2α−1k+α−1, and k∇uk3L3 ≤CkukB˙−α∞,∞kuk2
H˙3+α2
with −1< α <1, (1.30) kuk˙
H3+α2 ≤Ck∇uk
1−α 2
L2 k∆ukL1+α22 with −1< α <1, (1.31) and
k∇bk2L4 ≤Ck∇bkB˙∞,∞−β kbkH˙1+β with 0< β <1, (1.32) kbkH˙1+β ≤Ck∇bk1−βL2 k∆bkβL2 with 0< β <1. (1.33)
2. Proof of Theorem 1.1
Testing (1.1) by∂tdand using|d|= 1 andd·∂td= 0, we easily get the conser- vation of the energy:
d dt
Z
(|∂td|2+|∇d|2)dx= 0. (2.1) Applying the operator Λ1+s to equation (1.1), testing by Λ1+s∂td, using (1.18), (1.16), (1.17), (1.20), (1.21) and (1.22), we reach
1 2
d dt
Z
(|Λ1+s∂td|2+|Λ2+sd|2)dx
= Z
Λ1+s(d|∇d|2−d|∂td|2)Λ1+s∂tddx
≤(kΛ1+s(d|∇d|2)kL2+kΛ1+s(d|∂td|2)kL2)kΛ1+s∂tdkL2
≤C(kdkL∞kΛ1+s(|∇d|2)kL2+k∇dk2L2pkΛ1+sdk
L
2p
p−2)kΛ1+s∂tdkL2
+C(kdkL∞kΛ1+s(|∂td|2)kL2+k∂tdk2L2nkΛ1+sdk
Ln−22n )kΛ1+s∂tdkL2
≤C(k∇dkL∞kΛ2+sdkL2+k∇2dkLpkΛ1+sdk
L
2p
p−2)kΛ1+s∂tdkL2
+C(k∂tdkL∞kΛ1+s∂tdkL2+k∂tdkLnk∂tdkL∞kΛ2+sdkL2)kΛ1+s∂tdkL2
≤Ck∇dkL∞kΛ2+sdkL2kΛ1+s∂tdkL2
+Ck∂tdkL∞kΛ1+s∂tdk2L2+Ck∂tdkL∞kΛ2+sdkL2kΛ1+s∂tdkL2
≤C(k∇dkL∞+k∂tdkL∞)(kΛ2+sdk2L2+kΛ1+s∂tdk2L2)
≤C(1 +k∇dkB˙0∞,∞+k∂tdkB˙∞,∞0 ) log(e+y2)y2, withy2:=kΛ1+s∂tdk2L2+kΛ2+sdk2L2, which gives
sup
0≤t≤T
(kΛ1+s∂tdk2L2+kΛ2+sdk2L2)≤C.
This completes the proof.
3. Proof of Theorem 1.2
Since it is easy to prove that there are T0 > 0 and a unique strong solution (u, π, d) to the problem (1.4)-(1.7) in [0, T0], we only need to prove a priori estimates.
Testing (1.4) byuand using (1.6), we see that 1
2 d dt
Z
|u|2dx+ Z
|∇u|2dx=− Z
(u· ∇)d·∆ddx. (3.1) Testing (1.5) by−∆d, usingd∆d=−|∇d|2and|d|= 1, we find that
1 2
d dt
Z
|∇d|2dx+ Z
|∆d|2dx= Z
(u· ∇)d·∆ddx+ Z
(d∆d)2dx
≤ Z
(u· ∇)d·∆ddx+ Z
|∆d|2dx.
(3.2)
Summing up (3.1) and (3.2), we have Z
(|u|2+|∇d|2)dx≤ Z
(|u0|2+|∇d0|2)dx. (3.3)
ApplyingDsto (1.4), testing byDsu, using (1.6), (1.18), (1.19), (1.23), (1.24), (1.25), (1.26) and (1.27), we obtain
1 2
d dt
Z
|Dsu|2dx+ Z
|D1+su|2dx
=− Z
(Ds(u· ∇u)−u∇Dsu)Dsudx+ Z
Ds(∇d ∇d) :∇Dsudx
≤Ck∇ukLq1kDsuk
L
2q1
q1−2kDsukL2+Ck∇dkLq2kDs∇dk
L
2q2
q2−2k∇DsukL2
≤Ck∇ukB˙−α∞,∞kDs+αukL2kDsukL2+Ck∇dkB˙∞,∞−α kDs+α∇dkL2kDΛsukL2
≤Ck∇ukB˙−α∞,∞kDsuk2−αL2 kD1+sukαL2
+Ck∇2dk
1−α 2−α
B˙∞,∞−α kDs+1dk1−αL2 kDs+2dkαL2kD1+sukL2
≤ 1
8kD1+suk2L2+1
8kDs+2dk2L2+Ck∇uk
2 2−α
B˙∞,∞−α kDsuk2L2
+Ck∇2dk
2 2−α
B˙∞,∞−α kDs+1dk2L2.
(3.4)
ApplyingDs+1 to (1.5), testing byDs+1dand using (1.6), we obtain 1
2 d dt
Z
|Ds+1d|2dx+ Z
|Ds+2d|2dx
= Z
Ds+1(d|∇d|2)Ds+1ddx
− Z
(Ds+1(u· ∇d)−u∇Ds+1d)Ds+1d dx=:I1+I2.
(3.5)
Using (1.18),|d|= 1, (1.25), (1.26), and (1.27), we boundI1as follows.
I1≤ kDs+1(d|∇d|2)k
L
2q2 q2 +2
kDs+1dk
L
2q2 q2−2
≤C(kdkL∞kDs+1(|∇d|2)k
L
2q2 q2 +2
+k∇dk2Lq2kDs+1dk
L
2q2
q2−2)kDs+1dk
L
2q2 q2−2
≤C(k∇dkLq2kDs+2dkL2+k∇dk2Lq2kDs+1dk
L
2q2
q2−2)kDs+1dk
L
2q2 q2−2
≤Ck∇dk2Lq2kDs+1dk2
L
2q2 q2−2
+ 1
16kDs+2dk2L2
≤Ck∇dk2B˙−α
∞,∞k∇dk2H˙s+α+ 1
16kDs+2dk2L2
≤ 1
8kDs+2dk2L2+Ck∇2dk
2 2−α
B˙−α∞,∞kDs+1dk2L2. Using the Leibniz rule, we writeI2 as follows.
I2=− Z
(C1DuDs+1d+
s
X
k=2
CkDkuDs+2−kd+Cs+1Ds+1u· ∇d)Ds+1ddx
=:I21+
s
X
k=2
I2k+I2s+1.
(3.6)
By the same calculations as that ofI1, we have I2s+1≤Ck∇dkLq2kDs+1dk
L
2q2
q2−2kDs+1ukL2
≤ 1
16kDs+1uk2L2+Ck∇2dk
2 2−α
B˙∞,∞−α kDs+1dk2L2.
(3.7)
Using (1.23) and (1.24), we boundI21as follows.
I21≤Ck∇ukLq1kDs+1dk
L
2q1
q1−2kDs+1dkL2
≤Ck∇uk1−θ˙ 1
B∞,∞−α kukθ˙1
Hs+α· k∇2dkθ˙1
B∞,∞−α k∇dk1−θ˙ 1
Hs+αkDs+1dkL2
≤C(k∇ukB˙−α∞,∞+k∇2dkB˙∞,∞−α )(kukH˙s+α+k∇dkH˙s+α)kDs+1dkL2
≤ 1
16kDs+1uk2L2+ 1
16kDs+2dk2L2
+C(k∇uk
2 2−α
B˙∞,∞−α +k∇2dk
2 2−α
B˙−α∞,∞)(kDsuk2L2+kDs+1dk2L2).
(3.8)
Using (1.28) and (1.29), we boundPs
k=2I2k as follows.
s
X
k=2
I2k ≤Ck∇uk1−˙ θ˜k
B∞,∞−α kukθ˜˙k
Hs+αk∇2dkθ˜˙k
B−α∞,∞k∇dk1−˙ θ˜k
Hs+αkDs+1dkL2
≤C(k∇ukB˙∞,∞−α +k∇2dkB˙−α∞,∞)(kukH˙s+α+k∇dkH˙s+α)kDs+1dkL2
≤C(k∇ukB˙∞,∞−α +k∇2dkB˙−α∞,∞)(kukH˙s+α+k∇dkH˙s+α)
×(kDsukL2+kDs+1dkL2)
≤ 1
16kDs+1uk2L2+ 1
16kDs+2dk2L2
+C(k∇uk
2 2−α
B˙−α∞,∞+k∇2dk
2 2−α
B˙−α∞,∞)(kDsuk2L2+kDs+1dk2L2).
(3.9)
Inserting the above estimates into (3.5) and combining with (3.4) and using the Gronwall inequality, we arrive at
kDsukL∞(0,T;L2)+kDs+1dkL∞(0,T;L2)≤C.
This completes the proof.
4. Proof of Theorem 1.3
We only need to show a priori estimates. For simplicity, we will takeh=γ= 1.
First, testing (1.10) byuand using (1.12), we see that 1
2 d dt
Z
|u|2dx+ Z
|∇u|2dx= Z
(b· ∇)b·udx. (4.1) Testing (1.11) byband using (1.12), we find that
1 2
d dt
Z
|b|2dx+ Z
|∇b|2dx+ Z
|b×rotb|2dx= Z
(b· ∇)u·bdx. (4.2) Summing up (4.1) and (4.2), we obtain
1 2
d dt
Z
(|u|2+|b|2)dx+ Z
(|∇u|2+|∇b|2+|b×rotb|2)dx= 0. (4.3)
Testing (1.10) by−∆u, using (1.12), (1.30) and (1.31), we infer that 1
2 d dt
Z
|∇u|2dx+ Z
|∆u|2dx
= Z
(u.∇)u·∆udx− Z
(b· ∇)b·∆udx
=−X
i,j
Z
∂jui∂iu∂judx− Z
(b· ∇)b·∆udx
≤Ck∇uk3L3+kbkL∞k∇bkL2k∆ukL2
≤CkukB˙∞,∞−α kuk2
H˙ 3+α2
+Ck∇bkL2k∆ukL2
≤CkukB˙∞,∞−α k∇uk1−αL2 k∆uk1+αL2 +Ck∇bkL2k∆ukL2
≤ 1
8k∆uk2L2+Ckuk1−α˙2
B∞,∞−α
k∇uk2L2+Ck∇bk2L2.
(4.4)
Testing (1.11) by−∆band using (1.12), we deduce that 1
2 d dt
Z
|∇b|2dx+ Z
|∆b|2dx
= Z
(u· ∇)b·∆bdx− Z
(b· ∇)u·∆bdx +
Z
(rotb×b) rot ∆bdx− Z
[(rotb×b)×b] rot ∆bdx
=:`1+`2+`3+`4.
(4.5)
We bound`1 and`2 as follows.
`1=X
i,j
Z
ui∂ib∂j2bdx=−X
i,j
Z
∂jui∂ib∂jbdx≤Ck∇ukL2k∇bk2L4
≤Ck∇ukL2kbkL∞k∆bkL2≤Ck∇ukL2k∆bkL2≤ 1
16k∆bk2L2+Ck∇uk2L2.
`2≤ kbkL∞k∇ukL2k∆bkL2 ≤Ck∇ukL2k∆bkL2 ≤ 1
16k∆bk2L2+Ck∇uk2L2. Using (1.32) and (1.33), we bound`3 and`4 as follows.
`3=−X
i
Z
(rotb×∂ib)∂irotbdx≤Ck∇bk2L4k∆bkL2
≤Ck∇bkB˙−β
∞,∞kbkH˙1+βk∆bkL2 ≤Ck∇bkB˙−β
∞,∞k∇bk1−βL2 k∆bk1+βL2
≤ 1
16k∆bk2L2+Ck∇bk
2 1−β
B˙−β∞,∞
k∇bk2L2.
`4=X
i
Z
∂i[(rotb×b)×b]∂irotbdx≤X
i
Z
[(rotb×∂ib)×b]∂irotbdx
+X
i
Z
[(rotb×b)×∂ib]∂irotbdx≤CkbkL∞k∇bk2L4k∆bkL2
≤ 1
16k∆bk2L2+Ck∇bk
2 1−β
B˙∞,∞−β k∇bk2L2.
Inserting the above estimates into (4.5), and combining this with (4.4), and using the Gronwall inequality, we conclude that
k∇ukL∞(0,T;L2)+k∇bkL∞(0,T;L2)≤C. (4.6) This completes the proof by (1.15).
Acknowledgment. This work is partially supported by NSFC (No. 11171154).
The authors would like to thank the referees for their careful reading and helpful suggestions.
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Jishan Fan
Department of Applied Mathematics, Nanjing Forestry University, Nanjing 210037, China
E-mail address:[email protected]
Yong Zhou (corresponding author)
School of Mathematics, Shanghai University of Finance and Economics, Shanghai 200433, China.
Department of Mathematics, King Abdulaziz University, Jeddah 21589, Saudi Arabia E-mail address:[email protected]