Global regularity for the 2D MHD equations with mixed partial dissipation and magnetic diffusion

0.16 MB·English

by  Chongsheng Cao

GLOBAL REGULARITY FOR THE 2D MHD EQUATIONS WITH MIXED PARTIAL
DISSIPATION AND MAGNETIC DIFFUSION
9
0 CHONGSHENGCAOANDJIAHONGWU
0
2
Abstract. Whether or not classical solutions of the 2D incompressible MHD equations without full

n dissipationandmagneticdiffusioncandevelopfinite-timesingularitiesisadifficultissue. Amajorresult
a of this paper establishes the global regularity of classical solutions for the MHD equations with mixed
J partial dissipation and magnetic diffusion. In addition, the global existence, conditional regularity and
9 uniqueness ofaweaksolutionisobtainedforthe2DMHDequations withonlymagneticdiffusion.
1

]
P
1. Introduction
A
. This paper concerns itself with the fundamental issue of whether classical solutions of the 2D incom-
h
pressibleMHDequationscandevelopfinite-timesingularities. The2DMHDequationsunderconsideration
t
a assume the form
m
u +u u= p+ν u +ν u +b b, (1)
[ t ·∇ −∇ 1 xx 2 yy ·∇

1 bt+u·∇b=η1 bxx+η2 byy+b·∇u, (2)
v u=0, (3)
∇·
8
b=0, (4)
0 ∇·
9 where (x,y) R2, t 0, u = (u (x,y,t), u (x,y,t)) denotes the 2D velocity field, p = p(x,y,t) denotes
1 2
2 the pressure,∈b = (b (≥x,y,t),b (x,y,t)) denotes the magnetic field, and ν , ν , η and η are nonnegative
1 2 1 2 1 2
.
1 real parameters.
0
9 When ν1 >0, ν2 >0, η1 > 0 and η2 >0, (1)-(4) has a unique global classical solution for every initial
0 data(u ,b ) Hm withm 2(seee.g. [4],[7]). However,ifanyoneoftheseparametersiszero,theglobal
0 0
: regularity iss∈ue has not bee≥n settled. This paper establishes the global regularity of classical solutions of
v
i (1)-(4) with either ν1 = 0, ν2 = ν > 0, η1 = η > 0 and η2 = 0 or ν1 = ν > 0, ν2 = 0, η1 = 0 and
X
η =η >0. More precisely, we have the following theorem.
2
r
a Theorem 1. Consider the 2D MHD equations (1)-(4) with ν = 0, ν = ν > 0, η = η > 0 and η = 0.
1 2 1 2
Assume u H2(R2) and b H2(R2) with u =0 and b =0. Then (1)-(4) with the initial data
0 0 0 0
∈ ∈ ∇· ∇·
(u ,b ) has a unique global classical solution (u,b). In addition, (u,b) satisfies
0 0
(u,b) L∞([0, );H2), ω L2([0, );H1), j L2([0, );H1), (5)
y x
∈ ∞ ∈ ∞ ∈ ∞
where ω = u and j = b represent the vorticity and the current density, respectively.
∇× ∇×
A similar global regularity result can also be stated for (1)-(4) with ν = ν > 0, ν = 0, η = 0 and
1 2 1
η =η >0.
2
Attentionisalsopaidtothe2DMHDequationswithoutdissipationbutwithmagneticdiffusion,namely
(1)-(4) with ν = ν = 0 but with η = η = η > 0. In this case, we obtain the following global a priori
1 2 1 2
2000 Mathematics Subject Classification. 35B45,35B65,76W05.
Key words and phrases. classicalsolutions,globalregularity,MHDequations, partialdissipationandmagneticdiffusion.
1
2 CHONGSHENGCAOANDJIAHONGWU
bound for ω = u and j = b,
∇× ∇×
t
ω(t) 2+ j(t) 2+η j(τ) 2dτ C(η)( ω(0) 2+ j(0) 2) for t 0,
k k2 k k2 k∇ k2 ≤ k k2 k k2 ≥
Z0
where C(η) is a constant depending on η only. One consequence of this global bound is the existence of
a global H1-weak solution. It is not clear if such weak solutions are unique or can be improved to global
classical solutions. However, if we know the velocity field u of a solution obeys
1 T
sup u(t) dt< , (6)
p
p≥2 √p Z0 k∇ k ∞
then this solution actually becomes a classical solution on [0,T] and two weak solutions with one of their
velocities satisfying this bound must coincide on [0,T]. We remark that (6) is weaker than the standard
T
condition u(t) dt< and, as some preliminary evidence shows,is more likely to be proventrue
0 k∇ k∞ ∞
for (1)-(4) with η =η =η >0.
1 2
R
This work is partially motivated by the recent progress made by Chae [2], Hou and Li [5] and Danchin
and Paicu [3] on the 2D Boussinesq equations,
u +u u= p+ν∆u+θe , (7)
t 2
·∇ −∇
u=0, (8)
∇·
θ +u θ =η∆θ, (9)
t
·∇
where the 2D vector u represents the velocity field, the scalar θ the temperature, and e = (0,1). Chae
2
[2] and Hou and Li [5] independently established the globalregularity of (7)-(8) with either dissipation or
thermaldiffusion. DanchinandPaicu[3]constructedglobalsolutionsof(7)-(8)witheitherη =0andν∆u
replacedby νu or ν =0 and η∆θ by ηθ . We remarkthat the globalregularityissue for the 2D MHD
xx xx
equations (1)-(4) is more sophisticated. The equations of u and b in (1)-(4) are both nonlinearly coupled
vectors equations and the approaches in [2],[3] and [5] do not appear to apply. In fact, it is not clear if
(1)-(4) with η =η =0 or (1)-(4) with ν =η =0 has global classical solutions.
1 2 2 2
Therestofthispaperisdividedintotwosections. Thesecondsectionisdevotedtotheglobalregularity
of (1)-(4) with either ν = 0, ν = ν > 0, η = η > 0 and η = 0 or ν = ν > 0, ν = 0, η = 0 and
1 2 1 2 1 2 1
η = η > 0. The third section handles (1)-(4) with ν = ν = 0 and η = η = η > 0. Throughout these
2 1 2 1 2
sections the Lp-norm of a function f is denoted by f p, the Hs-norm by f Hs and the norm in the
k k k k
Sobolev space Ws,p by f Ws,p.
k k
2. Mixed partial dissipation and magnetic diffusion
This section proves Theorem 1 as well as a parallel result for the case when ν =ν >0, ν =0, η =0
1 2 1
and η = η > 0. The proof of Theorem 1 is achieved through two stages. The first stage establishes a
2
global bound for ω(t) and j(t) while the second obtains a bound for ω(t) and j(t) . The
2 2 2 2
k k k k k∇ k k∇ k
following elementary lemma will play an important role.
2.1. An elementary lemma.
Lemma 1. Assume that f, g, g , h and h are all in L2(R2). Then,
y x
fgh dxdy C f g 1/2 g 1/2 h 1/2 h 1/2. (10)
| | ≤ k k2 k k2 k yk2 k k2 k xk2
ZZ
2D MHD WITH MIXED PARTIAL DISSIPATION AND MAGNETIC DIFFUSION 3
Proof. Applying Ho¨lder’s inequality and the elementary inequality
1 1
4 4
sup F(x) √2 F(x)2dx F (x)2dx , (11)
x
x∈R| |≤ (cid:18)Z | | (cid:19) (cid:18)Z | | (cid:19)
we have
fgh dxdy
| |
ZZ
1/2 1/2
C f 2 dx g 2 dx sup h dy
≤ Z "(cid:18)Z | | (cid:19) (cid:18)Z | | (cid:19) (cid:18)−∞<x<∞ (cid:19)#
1/2 1/2 1/4 1/4
C f 2 dx g 2 dx h2 dx h 2 dx dy
x
≤ Z "(cid:18)Z | | (cid:19) (cid:18)Z | | (cid:19) (cid:18)Z | | (cid:19) (cid:18)Z | | (cid:19) #
1/2
1 1
C f sup g 2 dx h 2 h 2. (12)
≤ k k2 −∞<y<∞(cid:18)Z | | (cid:19) ! k k2 k xk2
In addition, by (11) again,
4
sup g 2 dx
| |
−∞<y<∞(cid:18)Z (cid:19)
2 2
C g 2 dx dy g g dx dy
y
≤ "Z (cid:18)Z | | (cid:19) # "Z (cid:18)Z | || | (cid:19) #
1/2 2
C g 4 dy dx g 2 dx g 2 dx dy
y
≤ Z (cid:18)Z | | (cid:19) ! Z (cid:20)(cid:18)Z | | (cid:19)(cid:18)Z | | (cid:19)(cid:21)
3/4 1/4 2
C g 2 dy g 2 dy dx
y
≤ Z "(cid:18)Z | | (cid:19) (cid:18)Z | | (cid:19) # !
sup g 2 dx g 2 dxdy
y
× | | | |
(cid:18)−∞<y<∞Z (cid:19) (cid:18)ZZ (cid:19)
C g 3 g sup g 2 dx g 2.
≤ k k2 k yk2 | | k yk2
(cid:18)−∞<y<∞Z (cid:19)
That is,
sup g 2 dx C g g . (13)
2 y 2
| | ≤ k k k k
−∞<y<∞
Z
Combining (12) and (13) yields (10). This completes the proof of Lemma 1. (cid:3)
2.2. A priori bounds for ω and j . This subsection establishes a priori bounds for ω and j
2 2 2 2
k k k k k k k k
as stated in the following proposition.
Proposition 2. If (u,b) solves (1)-(4) with ν =0, ν =ν >0, η =η >0 and η =0, then the vorticity
1 2 1 2
ω = u and the current density j = b satisfy
∇× ∇×
t t
ω(t) 2+ j(t) 2+ν ω (τ) 2 dτ +η j (τ) 2 dτ C(ν,η) ω 2+ j 2 (14)
k k2 k k2 k y k2 k x k2 ≤ k 0k2 k 0k2
Z0 Z0
(cid:0) (cid:1)
where C(ν,η) denotes a constant depending on ν and η only, ω = u and j = b .
0 0 0 0
∇× ∇×
4 CHONGSHENGCAOANDJIAHONGWU
Proof. Takingthe innerproductsof(1)withuand(2)withb,addingtheresultsandintegratingbyparts,
we obtain
t t
u(t) 2+ b(t) 2+2ν u (τ) 2dτ +2η b (τ) 2dτ u 2+ b 2. (15)
k k2 k k2 k y k2 k x k2 ≤k 0k2 k 0k2
Z0 Z0
Since ω and j satisfy
ω +u ω =ν ω +b j, (16)
t yy
·∇ ·∇
j +u j =ηj +b ω+2∂ b (∂ u +∂ u ) 2∂ u (∂ b +∂ b ), (17)
t xx x 1 x 2 y 1 x 1 x 2 y 1
·∇ ·∇ −
we find that X(t)= ω(t) 2+ j(t) 2 obeys
k k2 k k2
1 dX(t)
+ν ω 2+η j 2 2 [∂ b (∂ u +∂ u ) ∂ u (∂ b +∂ b )]j dxdy .
2 dt k yk2 k xk2 ≤ x 1 x 2 y 1 − x 1 x 2 y 1
(cid:12)Z (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Applying Lemma 1, we can bound the(cid:12)terms on the right as follows. C’s in these estimate(cid:12)s denote either
pure constants or constants depending on ν and η only.
∂ b ∂ u j dxdy
x 1 x 2
| || || |
Z
C ∂ u 1/2 ∂ u 1/2 j 1/2 j 1/2 ∂ b
≤ k x 2k2 k xy 2k2 k k2 k xk2 k x 1k2
ν η
∂ u 2+ j 2+C ∂ u ∂ b 2 j
≤ 4 k xy 2k2 8k xk2 k x 2k2k x 1k k k2
ν η
ω 2+ j 2+C ω ∂ b 2 j
≤ 4 k yk2 8k xk2 k k2 k x 1k2 k k2
ν η
ω 2+ j 2+C ∂ b 2X(t).
≤ 4 k yk2 8k xk2 k x 1k2
∂ b ∂ u j dxdy
x 1 y 1
| || || |
Z
1 1 1 1
C ∂ b 2 ∂ b 2 ∂ u 2 ∂ u 2 j
≤ k x 1k2 k xx 1k2 k y 1k2 k yy 1k2 k k
ν η
∂ u 2+ ∂ b 2+C ∂ b ∂ u j 2
≤ 4 k yy 1k2 8k xx 1k2 k x 1kk y 1k2k k2
ν η
ω 2+ j 2+C( ∂ b 2+ ∂ u 2) j 2.
≤ 4 k yk2 8k xk2 k x 1k2 k y 1k2 k k2
∂ u ∂ b jdxdy
x 1 x 2
(cid:12)Z (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) = (u1∂xxb2(cid:12)j+u1∂xb2jx)dxdy
(cid:12)Z (cid:12)
(cid:12) 1 1 1 1 (cid:12) 1 1 1 1
C(cid:12) u 2 ∂ u 2 j 2 j 2 ∂(cid:12) b +C u 2 ∂ u 2 ∂ b 2 ∂ b 2 j
≤(cid:12) k 1k2 k y 1k2 k k2 k xk2 k x(cid:12)x 2k2 k 1k2 k y 1k2 k x 2k2 k xx 2k2 k xk2
1 1 1 3 1 1 1 3
C u 2 ∂ u 2 j 2 j 2 +C u 2 ∂ u 2 ∂ b 2 j 2
≤ k 1k2 k y 1k2 k k2 k xk2 k 1k2 k y 1k2 k x 2k2 k xk2
η
j 2+C u 2 ∂ u 2 j 2+C u 2 ∂ u 2 ∂ b 2
≤ 8k xk2 k 1k2 k y 1k2 k k2 k 1k2 k y 1k2 k x 2k2
η
j 2+C u 2 ∂ u 2 j 2.
≤ 8k xk2 k 1k2 k y 1k2 k k2
2D MHD WITH MIXED PARTIAL DISSIPATION AND MAGNETIC DIFFUSION 5
∂ u ∂ b jdxdy
x 1 y 1
(cid:12)Z (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (u1∂xyb1(cid:12)j+u1∂yb1jx)dxdy
≤
(cid:12)Z (cid:12)
(cid:12) 1 1 1 1 (cid:12) 1 1 1 1
C(cid:12) u 2 ∂ u 2 j 2 j 2 ∂(cid:12) b +C u 2 ∂ u 2 ∂ b 2 ∂ b 2 j
≤(cid:12) k 1k2 k y 1k2 k k2 k xk2 k (cid:12)xy 1k2 k 1k2 k y 1k2 k y 1k2 k xy 1k2 k xk2
1 1 1 3 1 1 1 3
C u 2 ∂ u 2 j 2 j 2 +C u 2 ∂ u 2 ∂ b 2 j 2
≤ k 1k2 k y 1k2 k k2 k xk2 k 1k2 k y 1k2 k y 1k2 k xk2
η
j 2+C u 2 ∂ u 2 j 2+C u 2 ∂ u 2 ∂ b 2
≤ 8 k xk2 k 1k2 k y 1k2 k k2 k 1k2 k y 1k2 k y 1k2
η
j 2+C u 2 ∂ u 2 j 2.
≤ 8k xk2 k 1k2 k y 1k2 k k2
Combining these estimates, we have
dX(t)
+ν ω 2+η j 2 C( ∂ u 2+ ∂ b 2)X(t),
dt k yk2 k xk2 ≤ k y 1k2 k x 1k2
which, together with (15), yields (14). (cid:3)
2.3. A priori bounds for ω and j . Thissubsectionprovidesglobala priori bounds for ω
2 2 2
k∇ k k∇ k k∇ k
and j .
2
k∇ k
Proposition 3. If (u,b) solves (1)-(4) with ν =0, ν =ν >0, η =η >0 and η =0, then the vorticity
1 2 1 2
ω and the current density j satisfy
t t
ω(t) 2+ j(t) 2+ν ω (τ) 2 dτ +η j (τ) 2 dτ C(ν,η) ω 2+ j 2 (18)
k∇ k2 k∇ k2 k∇ y k2 k∇ x k2 ≤ k∇ 0k2 k∇ 0k2
Z0 Z0
(cid:0) (cid:1)
where C(ν,η) denotes a constant depending on ν and η only.
Proof. Taking the inner products of (16) with ∆ω leads to
1 d
ω 2+ ν ω 2
2 dtk∇ k2 k∇ yk2
= ω u ω dxdy+ ω b j dxdy+ b ( j) ω dxdy.
− ∇ ·∇ ·∇ ∇ ·∇ ·∇ ·∇ ∇ ·∇
Z Z Z
Similarly, taking the inner product of (17) with ∆j yields
1 d
j 2+ η j 2
2 dtk∇ k2 k∇ xk2
= j u j dxdy+ j b ω dxdy+ b ( ω) j dxdy
− ∇ ·∇ ·∇ ∇ ·∇ ·∇ ·∇ ∇ ·∇
Z Z Z
+2 [∂ b (∂ u +∂ u )] j dxdy 2 [∂ u (∂ b +∂ b )] j dxdy.
x 1 x 2 y 1 x 1 x 2 y 1
∇ ·∇ − ∇ ·∇
Z Z
Adding the above equations and integrating by parts, we find
1 d
( ω 2+ j 2) + ν ω 2+η j 2 =I +I +I +I +I ,
2 dt k∇ k2 k∇ k2 k∇ yk2 k∇ xk2 1 2 3 4 5
6 CHONGSHENGCAOANDJIAHONGWU
where
I = ω u ω dxdy,
1
− ∇ ·∇ ·∇
Z
I = j u j dxdy,
2
− ∇ ·∇ ·∇
Z
I =2 ω b j dxdy,
3
∇ ·∇ ·∇
Z
I =2 [∂ b (∂ u +∂ u )] j dxdy,
4 x 1 x 2 y 1
∇ ·∇
Z
I = 2 [∂ u (∂ b +∂ b )] j dxdy.
5 x 1 x 2 y 1
− ∇ ·∇
Z
To bound I , we write the integrand explicitly and further divide it into four terms
1
I = (∂ u ω2+∂ u ω ω +∂ u ω ω +∂ u ω2)dxdy
1 x 1 x x 2 x y y 1 x y y 2 y
Z
= I +I +I +I .
11 12 13 14
By the divergence-free condition ∂ u +∂ u =0 and Lemma 1,
x 1 y 2
I = ∂ u ω2 dxdy
11 − y 2 x
Z
1 1 1 1
C ∂ u 2 ∂ u 2 ω 2 ω 2 ω
≤ k y 2k2 k xy 2k2 k xk2 k xyk2 k xk
1 1 1 3
C ω 2 ω 2 ω 2 ω 2
≤ k k2 k yk2 k∇ yk2 k∇ k2
ν 2 2
ω 2+C ω 3 ω 3 ω 2.
≤ 10 k∇ yk2 k k2 k yk2 k∇ k2
By Lemma 1,
1 1 1 1
I C ∂ u 2 ∂ u 2 ∂ ω 2 ∂ ω 2 ω
12 ≤ k x 2k2 k xy 2k2 k y k2 k xy k2 k xk2
1 1 1 3
C ω 2 ω 2 ω 2 ω 2
≤ k k2 k yk2 k∇ yk2 k∇ k2
ν 2 2
ω 2+C ω 3 ω 3 ω 2.
≤ 10k∇ yk2 k k2 k yk2 k∇ k2
I and I can be similarly bounded,
13 14
ν 2 2
I , I ω 2+C ω 3 ω 3 ω 2.
13 14 ≤ 10 k∇ yk2 k k2 k yk2 k∇ k2
I and I can be bounded by applying Lemma 1.
2 3
1 1 1 1
I C u 2 u 2 j 2 j 2 j
2 ≤ k∇ k2 k∇ yk2 k∇ k2 k∇ xk2 k∇ k2
1 1 3 1
C ω 2 ω 2 j 2 j 2
≤ k k2 k yk2 k∇ k2 k∇ xk2
η 2 2
j 2+C ω 3 ω 3 j 2.
≤ 16 k∇ xk2 k k2 k yk2 k∇ k2
1 1 1 1
I C b ω 2 ω 2 j 2 j 2
3 ≤ k∇ k2 k∇ k2 k∇ yk2 k∇ k2 k∇ xk2
1 1 1 1
C j ω 2 ω 2 j 2 j 2
≤ k k2 k∇ k2 k∇ yk2 k∇ k2 k∇ xk2
ν η
ω 2+ j 2+C j 2 ω j
≤ 10 k∇ yk2 16 k∇ xk2 k k2 k∇ k2 k∇ k2
ν η
ω 2+ j 2+C j 2 ( ω 2+ j 2).
≤ 10 k∇ yk2 16 k∇ xk2 k k2 k∇ k2 k∇ k2
2D MHD WITH MIXED PARTIAL DISSIPATION AND MAGNETIC DIFFUSION 7
To bound I , we split it into two parts:
4
I = 2 ∂ [∂ b (∂ u +∂ u )]j dxdy+2 ∂ [∂ b (∂ u +∂ u )]j dxdy
4 x x 1 x 2 y 1 x y x 1 x 2 y 1 y
Z Z
I +I .
41 42
≡
Integrating by parts in I and applying Lemma 1, we have
41
I = 2 ∂ b (∂ u +∂ u )j dxdy
41 x 1 x 2 y 1 xx
−
Z
1 1 1 1
C ∂ b 2 ∂ b 2 ∂ u 2 ∂ u 2 j
≤ k x 1k2 k xx 1k2 k x 2k2 k xy 2k2 k xxk2
1 1 1 1
+C ∂ b 2 ∂ b 2 ∂ u 2 ∂ u 2 j
k x 1k2 k xx 1k2 k y 1k2 k yy 1k2 k xxk2
1 1 1 1
C j 2 j 2 ω 2 ω 2 j
≤ k k2 k∇ k2 k k2 k yk2 k∇ xk2
η
j 2+C ω j ω j
≤ 16 k∇ xk2 k k2 k k2 k∇ k2 k∇ k2
η
j 2+C ω j ( ω 2+ j 2).
≤ 16 k∇ xk2 k k2 k k2 k∇ k2 k∇ k2
I can be further decomposed into two parts:
42
I = 2 ∂ b (∂ u +∂ u )]j dxdy+2 ∂ b (∂ u +∂ u )]j dxdy
42 xy 1 x 2 y 1 y x 1 xy 2 yy 1 y
Z Z
I +I
421 422
≡
and these two terms can be bounded as follows.
1 1 1 1
I C ∂ b ∂ u 2 ∂ u 2 j 2 j 2
421 ≤ k xy 1k2 k x 2k2 k xy 2k2 k yk2 k xyk2
1 1 1 1
+C ∂ b ∂ u 2 ∂ u 2 j 2 j 2
k xy 1k2 k y 1k2 k yy 1k2 k yk2 k xyk2
1 1 3 1
C ω 2 ω 2 j 2 j 2
≤ k k2 k yk2 k∇ k2 k∇ xk2
η 2 2
j 2+C ω 3 ω 3 j 2
≤ 16 k∇ xk2 k k2 k yk2 k∇ k2
1 1 1 1
I C ∂ b 2 ∂ b 2 ∂ u j 2 j 2
422 ≤ k x 1k2 k xy 1k2 k xy 2k2 k yk2 k xyk2
1 1 1 1
+C ∂ b 2 ∂ b 2 ∂ u j 2 j 2
k x 1k2 k xy 1k2 k yy 1k2 k yk2 k xyk2
1 1 1 1
C j 2 j 2 ω j 2 j 2
≤ k k2 k xk2 k yk2 k yk2 k xyk2
η 2 2 4 2
j 2+C j 3 j 3 ω 3 j 3
≤ 16 k∇ xk2 k k2 k xk2 k∇ k2k∇ k2
η 2 2
j 2+C j 3 j 3 ( ω 2+ j 2).
≤ 16 k∇ xk2 k k2 k xk2 k∇ k2 k∇ k2
To bound I , we first write it into three terms,
5
I = 2 ∂ [∂ u (∂ b +∂ b )] j dxdy 2 ∂ [∂ u (∂ b +∂ b )] j dxdy
5 x x 1 x 2 y 1 x y x 1 x 2 y 1 y
− −
Z Z
= 2 ∂ u (∂ b +∂ b )j dxdy 2 ∂ u (∂ b +∂ b )j dxdy
x 1 x 2 y 1 xx xy 1 x 2 y 1 y
−
Z Z
2 ∂ u (∂ b +∂ b )j dxdy
x 1 xy 2 yy 1 y
−
Z
I +I +I .
51 52 53
≡
8 CHONGSHENGCAOANDJIAHONGWU
We bound these terms as follows.
1 1 1 1
I C ∂ u 2 ∂ u 2 ∂ b 2 ∂ b 2 j
51 ≤ k x 1k2 k xy 1k2 k x 2k2 k xx 2k2 k xxk2
1 1 1 1
+C ∂ u 2 ∂ u 2 ∂ b 2 ∂ b 2 j
k x 1k2 k xy 1k2 k y 1k2 k xy 1k2 k xxk2
1 1 1 1
C ω 2 ω 2 j 2 j 2 j
≤ k k2 k∇ k2 k k2 k∇ k2 k∇ xk2
η
j 2+C ω j ( ω 2+ j 2).
≤ 16 k∇ xk2 k k2 k k2 k∇ k2 k∇ k2
1 1 1 1
I C ∂ u ∂ b 2 ∂ b 2 j 2 j 2
52 ≤ k xy 1k2 k x 2k2 k xy 2k2 k yk2 k xyk2
1 1 1 1
+C ∂ u ∂ b 2 ∂ b 2 j 2 j 2
k xy 1k2 k y 1k2 k yy 1k2 k yk2 k xyk2
1 1 1 1
C ω 2 ω 2 j 2 j j 2
≤ k yk2 k∇ k2 k k2 k∇ k2 k∇ xk2
η 2 2 2 4
j 2+C ω 3 j 3 ω 3 j 3
≤ 16 k∇ xk2 k yk2 k k2 k∇ k2 k∇ k2
η 2 2
j 2+C ω 3 j 3 ( ω 2+ j 2).
≤ 16 k∇ xk2 k yk2 k k2 k∇ k2 k∇ k2
1 1 1 1
I C ∂ u 2 ∂ u 2 ∂ b j 2 j 2
53 ≤ k x 1k2 k xy 1k2 k xy 2k2 k yk2 k xyk2
1 1 1 1
+C ∂ u 2 ∂ u 2 ∂ b j 2 j 2
k x 1k2 k xy 1k2 k yy 1k2 k yk2 k xyk2
1 1 1 1
C ω 2 ω 2 j j 2 j 2
≤ k k2 k yk2 k∇ k2 k yk2 k xyk2
η 2 2
j 2+C ω 3 ω 3 j 2.
≤ 16 k∇ xk2 k k2 k yk2 k∇ k2
Collecting the above estimates, we finally obtain
d
( ω 2+ j 2) + ν ω 2+η j 2
dt k∇ k2 k∇ k2 k∇ yk2 k∇ xk2
2 2 2 2
C (( ω 3 + j 3)( ω 3 + j 3) + j ( ω + j ))( ω 2+ j 2).
≤ k yk2 k xk2 k k2 k k2 k k2 k k2 k k2 k∇ k2 k∇ k2
Applying the bound from Proposition 2, we find
t t
ω(t) 2+ j(t) 2+ν ω (τ) 2 dτ +η j (τ) 2 dτ C(ν,η)( ω 2+ j 2).
k∇ k2 k∇ k2 k∇ y k2 k∇ x k2 ≤ k∇ 0k2 k∇ 0k2
Z0 Z0
This completes the proof of Proposition 3. (cid:3)
2.4. Proof of Theorem 1. This subsection presents the proof of Theorem 1.
Proof of Theorem 1. With the a priori bounds of Propositions 2 and 3 at our disposal, the proof of this
theorem can be achieved through a parabolic regularization process. Let ǫ>0 be a small parameter and
consider a family of solutions (u ,b ) satisfying the regularized system of equations
ǫ ǫ
∂ u +u u = p +ν∂ u +b b +ǫ∆u , (19)
t ǫ ǫ ǫ ǫ yy ǫ ǫ ǫ ǫ
·∇ −∇ ·∇
∂ b +u b =η ∂ b +b u +ǫ∆b (20)
t ǫ ǫ ǫ xx ǫ ǫ ǫ ǫ
·∇ ·∇
u =0, (21)
ǫ
∇·
b =0, (22)
ǫ
∇·
u (x,0)=ψ u , b (x,0)=ψ b , (23)
ǫ ǫ 0 ǫ ǫ 0
∗ ∗
where ψ (x)=ǫ−2ψ(x/ǫ) with ψ satisfying
ǫ
ψ 0, ψ C∞(R2) and ψ =1.
≥ ∈ 0 k k1
2D MHD WITH MIXED PARTIAL DISSIPATION AND MAGNETIC DIFFUSION 9
Sinceu (x,0)andb (x,0)aresmooth,thestandardtheoryonthe2DviscousMHDequations(seee.g. [7])
ǫ ǫ
guaranteesthat (19)-(23) has a unique global smooth solution (u ,b ). It is easy to see that (u ,b ) obeys
ǫ ǫ ǫ ǫ
the a priori bounds in Propositions2 and 3 uniformly in ǫ. The solution (u, b) of (1)-(4) is then obtained
as a limit of (u , b ) and obey the bounds in Propositions 2 and 3.
ǫ ǫ
The uniqueness of the solutions follows from the elementary inequalities (see Lemma 14 of [3])
f C ( f + f + f ) and f C ( f + f + f ).
∞ 2 x 2 yy 2 ∞ 2 y 2 xx 2
k k ≤ k k k k k k k k ≤ k k k k k k
In fact, applying these inequalities, we have
t
( ω(τ) + j(τ) ) dτ
∞ ∞
k k k k
Z0
t
( ω(τ) + ω (τ) + ω (τ) ) dτ
2 y 2 y 2
≤ k k k k k∇ k
Z0
t
+ ( j(τ) + j (τ) + j (τ) ) dτ <
2 x 2 x 2
k k k k k∇ k ∞
Z0
for any t>0. It is well-known (see e.g. [1], [8]) that this bound yields the uniqueness. (cid:3)
2.5. (1)-(4) with ν = ν > 0, ν = 0, η = 0 and η = η > 0. A global regularity result similar to
1 2 1 2
Theorem 1 can be established for the 2D MHD equations (1)-(4) with ν = ν > 0, ν = 0, η = 0 and
1 2 1
η =η >0.
2
Theorem 4. Consider the 2D MHD equations (1)-(4) with ν = ν > 0, ν = 0, η = 0 and η = η > 0.
1 2 1 2
Assume u H2(R2) and b H2(R2) with u =0 and b =0. Then (1)-(4) has a unique global
0 0 0 0
∈ ∈ ∇· ∇·
classical solution (u,b). In addition, (u,b) satisfies
(u,b) L∞([0, );H2), ω L2([0, );H1), j L2([0, );H1), (24)
x y
∈ ∞ ∈ ∞ ∈ ∞
where ω = u and j = b represent the vorticity and the current density, respectively.
∇× ∇×
Proof. Although this theorem can be proven in a similar fashion as that of Theorem 1, we provide an
alternative proof. The idea is to convert (1)-(4) with ν = ν > 0, ν = 0, η = 0 and η = η > 0 into
1 2 1 2
a form dealt with by Theorem 1. Assume that (u,b) solves (1)-(4) with ν = ν > 0, ν = 0, η = 0 and
1 2 1
η =η >0. Set
2
U (x,y,t)=u (y,x,t), U (x,y,t)=u (y,x,t), P(x,y,t)=p(y,x,t),
1 2 2 1
B (x,y,t)=b (y,x,t), B (x,y,t)=b (y,x,t).
1 2 2 1
Then U =(U ,U ), P and B =(B ,B ) satisfy
1 2 1 2
U +U U = P +νU +B B, (25)
t yy
·∇ −∇ ·∇
B +U B =η B +B U, (26)
t xx
·∇ ·∇
U =0, (27)
∇·
B =0. (28)
∇·
The global regularity of (25)-(28) guaranteed by Theorem 1 allows us to obtain the global regularity for
(1)-(4) with ν =ν >0, ν =0, η =0 and η =η >0. This completes the proof of Theorem 4. (cid:3)
1 2 1 2
10 CHONGSHENGCAOANDJIAHONGWU
3. The MHD with magnetic diffusion
This section focuses on (1)-(4) with ν = ν = 0 and η = η = η > 0. Two major results are
1 2 1 2
established. Thefirstistheglobalexistenceofaweaksolutionandthesecondassessestheglobalregularity
and uniqueness of the weak solution under a suitable condition.
Theorem 5. Consider (1)-(4) with ν = ν = 0 and η = η = η > 0. Assume that (u ,b ) H1 with
1 2 1 2 0 0
∈
u =0 and b =0. Then (1)-(4) has a global weak solution (u,b) satisfying
0 0
∇· ∇·
u C([0, );H1), b C([0, );H1) L2([0, );H2). (29)
∈ ∞ ∈ ∞ ∩ ∞
The proof of this result relies on a global a priori bound for ω = u and j = b.
∇× ∇×
Theorem6. Assumetheinitialdata(u ,b ) H3, u =0and b =0. Let(u,b)bethecorresponding
0 0 0 0
∈ ∇· ∇·
solution of (1)-(4) with ν =ν =0 and η =η =η >0. If, for some T >0,
1 2 1 2
1 T
sup u(t) dt< , (30)
p
p≥2 √p Z0 k∇ k ∞
then (u,b) is regular on [0,T], namely
(u,b) C([0,T];H3).
∈
In addition, two weak solutions (u,b) and (u˜,˜b) in the regularity class (29) must be identical on the time
interval [0,T] if u satisfies (30).
The rest of this section is divided into four subsections. The first subsection presents a global a priori
bound for u H1 and b H1 and the second proves Theorem 5. The third subsection establishes a loga-
k k k k
rithmic Sobolev inequality, which serves as a preparation for the proof of Theorem 6. The last subsection
proves Theorem 6.
3.1. An a priori bound for u and b .
2 2
k∇ k k∇ k
Proposition 7. If (u,b) solves the 2D MHD equations (1)-(4) with ν = ν = 0 and η = η = η > 0,
1 2 1 2
then, for any t>0,
t
ω(t) 2+ j(t) 2+η j 2 dτ C(η)( u 2+ b 2), (31)
k k2 k k2 k∇ k2 ≤ k∇ 0k2 k∇ 0k2
Z0
where C(η) is a constant depending on η only. Therefore,
t
u(t) 2 + b(t) 2 +η b 2 dτ C(η)( u 2 + b 2 ). (32)
k kH1 k kH1 k kH2 ≤ k 0kH1 k 0kH1
Z0
Proof. It follows easily from (1) and (2) that, for any t>0,
t
u(t) 2+ b(t) 2+2η b(τ) 2 dτ = u(0) 2+ b(0) 2. (33)
k k2 k k2 k∇ k2 k k2 k k2
Z0
To prove (31), we employ the equations of the vorticity ω and the current density j,
ω +u ω =b j, (34)
t
·∇ ·∇
j +u j =η ∆j+b ω+2∂ b (∂ u +∂ u ) 2∂ u (∂ b +∂ b ). (35)
t x 1 x 2 y 1 x 1 x 2 y 1
·∇ ·∇ −
Taking the inner products of (34) with ω and of (35) with j, we find
1 d ω 2
k k2 = b jωdxdy,
2 dt ·∇
Z
1 d j 2
k k2 +η j 2 = b ωjdxdy+ 2 (∂ b (∂ u +∂ u ) ∂ u (∂ b +∂ b ))jdxdy.
2 dt k∇ k2 ·∇ x 1 x 2 y 1 − x 1 x 2 y 1
Z Z