T. K O B A Y A S H I (Tsukuba)
W. M. Z A J A ¸ C Z K O W S K I (Warszawa)
ON GLOBAL MOTION OF A COMPRESSIBLE BAROTROPIC VISCOUS FLUID WITH
BOUNDARY SLIP CONDITION
Abstract. Global-in-time existence of solutions for equations of viscous compressible barotropic fluid in a bounded domain Ω ⊂ R
3with the bound- ary slip condition is proved. The solution is close to an equilibrium solution.
The proof is based on the energy method. Moreover, in the L
2-approach the result is sharp (the regularity of the solution cannot be decreased) be- cause the velocity belongs to H
2+α,1+α/2(Ω × R
+) and the density belongs to H
1+α,1/2+α/2(Ω × R
+), α ∈ (1/2, 1).
1. Introduction. In this paper we prove the existence of global-in-time solutions to the following problem (see [5, 10]):
(1.1)
̺(v
t+ v · ∇v) − div T(v, p) = ̺f in Ω
T= Ω × (0, T ),
̺
t+ div(̺v) = 0 in Ω
T,
̺|
t=0= ̺
0, v|
t=0= v
0in Ω,
n · T(v, p) · τ
α+ γv · τ
α= 0, α = 1, 2, on S
T= S × (0, T ),
v · n = 0 on S
T,
where v = v(x, t) is the velocity of the fluid, ̺ = ̺(x, t) the density, f = f (x, t) the external force field, n the unit outward vector normal to S, τ
α, α = 1, 2, tangent vectors to S, 0 < γ the constant slip coefficient, and S = ∂Ω, where Ω ⊂ R
3is a bounded domain. By T(v, p) we denote the stress tensor of the form
(1.2) T (v, p) = {µ(∂
xiv
j+ ∂
xjv
i) + (ν − µ)δ
ijdiv v − δ
ijp}
i,j=1,2,3,
1991 Mathematics Subject Classification: 76N10, 35B35, 35G30.
Key words and phrases: global existence, compressible barotropic viscous fluid, bound- ary slip condition, Hilbert–Besov spaces.
[159]
where µ, ν are constant viscosity coefficients satisfying the thermodynamic restrictions
(1.3) ν >
13µ > 0.
Moreover, we consider the barotropic motion,
(1.4) p = p(̺).
From (1.1)
2,5it follows that the total mass of the fluid in Ω is conserved, (1.5)
\
Ω
̺ dx = M =
\
Ω
̺
0dx.
In this paper we prove existence of global-in-time solutions which are close to the equilibrium solution
(1.6) v
e= 0, ̺
e= M
|Ω| , where |Ω| = vol Ω.
The proof is based on a local existence result from [3] and on the prolon- gation technique from [12]. To recall the result from [3] we have to introduce the Lagrangian coordinates which are initial data to the Cauchy problem
(1.7) ∂x
∂t = v(x, t), x|
t=0= ξ ∈ Ω.
Integrating (1.7) we obtain a relation between the Eulerian x and the La- grangian ξ coordinates:
(1.8) x = ξ +
t
\
0
u(ξ, τ ) dτ ≡ x
u(ξ, t) ≡ x(ξ, t),
where u(ξ, t) = v(x(ξ, t), t). Moreover, we introduce η(ξ, t) = ̺(x(ξ, t), t), q(ξ, t) = p(η(ξ, t)), and g(ξ, t) = f (x(ξ, t), t).
Theorem 1.1. Assume that g ∈ H
α,α/2(Ω
T), v
0∈ H
1+α(Ω), ̺
0∈ H
1+α(Ω), 1/̺
0∈ L
∞(Ω), S ∈ C
2+α, α ∈ (1/2, 1). Then there exists a time T > 0 such that for t ≤ T there exists a solution (u, η) to problem (1.1) with u ∈ H
2+α,1+α/2(Ω
t), η ∈ H
1+α,1/2+α/2(Ω
t), and the following estimate holds:
kuk
2+α,Ωt+ kηk
1+α,Ωt≤ ϕ
1(T, kSk
C2+α, γ(T ))[kv
0k
1+α,Ω(1.9)
+ k̺
0k
1+α,Ω+ kgk
α,Ωt], t ≤ T,
|1/η|
∞,Ωt≤ ϕ
2(T, kSk
C2+α, γ(T ))|1/̺
0|
∞,Ω,
where γ(t) = kv
0k
1+α,Ω+ k̺
0k
1+α,Ω+ kgk
α,Ωtand ϕ
1, ϕ
2are increasing positive functions of their arguments.
To prove global existence we have to control the variation of the solu-
tion in a neighbourhood of the equilibrium solution. For this purpose we
introduce
(1.10) ̺
σ= ̺ − ̺
e, p
σ= p − p
e, where p
e= p(̺
e). Then (1.1) implies
(1.11)
̺(v
t+ v · ∇v) − div T(v, p
σ) = ̺f in Ω
T,
̺
σt+ v · ∇̺
σ+ ̺ div v = 0 in Ω
T, n · D(v) · τ
α+ γv · τ
α= 0, α = 1, 2, on S
T,
v · n = 0 on S
T,
v|
t=0= v
0, ̺
σ|
t=0= ̺
σ0≡ ̺
0− ̺
ein Ω, where
D (v) = {µ(v
ixj+ v
jxi) + (ν − µ)δ
ijdiv v}
i,j=1,2,3.
Sometimes instead of (1.11)
2we use the following equation for the pressure:
(1.12) p
σt+ v · ∇p
σ+ p
̺̺ div v = 0.
Applying Theorem 1.1 to problem (1.11) yields
Theorem 1.2. Let the assumptions of Theorem 1.1 hold. Let ̺
σ0∈ H
1+α(Ω). Then for solutions of the problem (1.11) we have the estimate
kuk
2+α,Ωt+ kη
σk
1+α,Ωt≤ ϕ
3(T, kSk
C2+α, γ(T ))[kv
0k
1+α,Ω(1.13)
+ k̺
σ0k
1+α,Ω+ kgk
α,Ωt], t ≤ T, where ϕ
3is an increasing positive function of its arguments and γ(T ) is defined in Theorem 1.1.
Since the local existence of solutions to problem (1.1) was proved in spaces such that u ∈ H
2+α,1+α/2(Ω
T), η ∈ H
1+α,1/2+α/2(Ω
T), we have to formulate problem (1.11) and (1.12) in Lagrangian coordinates. Hence we have
(1.14)
ηu
t− div
uT
u(u
,q
σ) = ηg in Ω
T, q
σt+ p
ηη div
uu = 0 in Ω
T, n
u· D
u(u) · τ
uα+ γu · τ
uα= 0, α = 1, 2, on S
T,
u · n
u= 0 on S
T,
u|
t=0= v
0, q
σ|
t=0= q
σ0≡ q
0− p
ein Ω, where q
σ= q − p
e, and
(1.15) η
t+ η div
uu = 0 in Ω
T, η|
t=0= v
0in Ω,
where ∇
u= ξ
ix∂
ξi, T
u(u, q
σ) = −q
σI + D
u(u), I is the unit matrix and
the operators D
u, div
uare obtained from D and div by replacing ∇ by ∇
u.
Finally n
u(ξ, t) = n(x(ξ, t)) and τ
u,α(ξ, t) = τ
α(x(ξ, t)), α = 1, 2.
Moreover, in view of the local solvability the Jacobi matrix with de- terminant J
x(ξ,t)of the transformation x = x(ξ, t), with elements a
ij= δ
ij+
Tt
0
∂
ξju
i(ξ, τ ) dτ , is invertible.
We have to underline that the use of the Lagrangian coordinates is not natural for problem (1.1) because the domain Ω is fixed. However in [3]
and also in this paper the Lagrangian coordinates have to be used because otherwise the nonlinear term v · ∇̺ in the continuity equation cannot be estimated in the proofs for v and ̺ such that v ∈ H
2+α,1+α/2(Ω
T), ̺ ∈ H
1+α,1/2+α/2(Ω
T), α ∈ (1/2, 1).
Moreover, we have not been able to show the invariance of the norm of H
2+α,1+α/2(Ω
T), α ∈ (1/2, 1), under the transformation (1.8), where u ∈ H
2+α,1+α/2(Ω
T). More precisely we examine the following problem. Let the transformation (1.8) be generated by w ∈ H
2+α,1+α/2(Ω
T), α ∈ (1/2, 1).
Then we have x = x
w(ξ, t). Let v = v(x, t) and u(ξ, t) = v(x
w(ξ, t), t). We want both u, v to belong to H
2+α,1+α/2(Ω
T). We can easily show that (1.16) c
1kvk
L2(0,T ;H2+α(Ω))≤ kuk
L2(0,T ;H2+α(Ω))≤ c
2kvk
L2(0,T ;H2+α(Ω)), where c
1, c
2are positive, nonvanishing and depend on kwk
2+α,ΩT.
However, we are not able to show (1.16) for v and u in H
2+α,1+α/2(Ω
T) under the assumption that c
1, c
2depend on kwk
2+α,ΩTonly. We can only show this if w is more regular but this is not appropriate for us.
Now we formulate the main result.
Theorem 1.3. Assume that the equilibrium solution is defined by (1.6) and the total mass of the fluid is determined by (1.5). Assume the thermo- dynamic restrictions (1.3) and that the barotropic motion with state equa- tion (1.4) is considered. Assume that the bounded domain Ω is not ro- tationally symmetric and S ∈ C
2+α, ̺
0, v
0∈ H
1+α(Ω), 1/̺
0∈ L
∞(Ω), g ∈ H
α,α/2(Ω
T), p ∈ C
2, α ∈ (1/2, 1), g = g(ξ, t) = f (x
u(ξ, t), t) and x = x
u(ξ, t) defined by (1.8) determines the transformation between the Eu- lerian and Lagrangian coordinates. Assume that kv
0k
1+α,Ω, k̺
0− ̺
ek
1+α,Ω, kgk
α,Ωt, t ∈ R
+are sufficiently small , where ̺
e= M/|Ω|. Then there exists a global solution to problem (1.1) such that
u ∈ H
2+α,1+α/2(Ω
t), η ∈ H
1+α,1/2+α/2(Ω
t),
t ∈ R
+, where u(ξ, t) = v(x
u(ξ, t), t), η(ξ, t) = ̺(x
u(ξ, t), t) and u deter- mines the transformation (1.8).
Finally, we want to add some historical remarks and comments concern-
ing global-in-time existence for viscous compressible fluids. The first results
concerning global-in-time existence for the Cauchy problem for equations
of viscous compressible and heat conducting fluids were obtained by Mat-
sumura and Nishida [6, 7]. The corresponding result for an initial boundary
value problem for the same equations was also shown by Matsumura and Nishida [8, 9]. They considered only a half space and they assumed that the initial density, velocity and temperature are from H
3and the external force has a gradient form. They assumed the Dirichlet boundary conditions for velocity and the Neumann boundary conditions for temperature. Valli [15] improved the results for the barotropic case showing global-in-time exis- tence for an arbitrary bounded domain Ω ⊂ R
3, general small external force field and for initial density and velocity from H
2(Ω). The result of [15] was extended to general viscous compressible heat-conducting fluid equations by Valli–Zaj¸aczkowski [16]. The global-in-time existence results from [6–9, 15, 16] follow from some a priori estimate whose proof depends heavily on the L
2-approach. However, the results are not sharp in the L
2-framework, where sharpness means that the existence result cannot be proved with less regular data. In the L
p-framework global-in-time existence of solutions for equations of viscous compressible heat-conducting fluids with Dirichlet boundary conditions was proved by Str¨ ohmer in [13, 14]. However his results are not sharp because he imposed too much regularity on data functions.
Our result is sharp for the L
2-approach and this is the reason why the fractional derivative spaces have been used. However this forced us to use the Lagrangian coordinates which are not appropriate for problems in fixed domains. Unfortunately in general our result is not sharp because then the L
p-approach is necessary.
Our result can be extended to the Dirichlet and Neumann problems and also to equations of viscous compressible heat-conducting fluids.
2. Notation and auxiliary results. First we introduce a partition of unity ({ e Ω
i}, {ζ
i}), Ω = S
i
Ω e
i. Let e Ω be one of the e Ω
i’s and ζ(ξ) = ζ
i(ξ) the corresponding function. If e Ω is an interior subdomain, then let e ω be such that e ω ⊂ e Ω and ζ(ξ) = 1 for ξ ∈ e ω. Otherwise we assume that e Ω ∩ S 6= ∅, e
ω ∩ S 6= ∅, e ω ⊂ e Ω. Let β ∈ e ω ∩ S ⊂ e Ω ∩ S, e S ≡ S ∩ e Ω. Introduce local coordinates {y} connected with {ξ} by
(2.1) y
k= α
kl(ξ
l− β
l), α
3k= n
k(β), k = 1, 2, 3,
where α
klis a constant orthogonal matrix, such that e S is determined by y
3= F (y
1, y
2), F ∈ H
3/2+αand
Ω = {y : |y e
i| < d, i = 1, 2, F (y
′) < y
3< F (y
′) + d, y
′= (y
1, y
2)}.
Next we introduce functions u
′and q
′by
(2.2) u
′i(y) = α
iju
j(ξ)|
ξ=ξ(y), q
′(y) = q(ξ)|
ξ=ξ(y),
where ξ = ξ(y) is the inverse transformation to (2.1). Further we introduce
new variables by
(2.3) z
i= y
i, i = 1, 2, z
3= y
3− e F (y), y ∈ e Ω,
which will be denoted by z = Φ(y), where e F is an extension of F to e Ω with F ∈ H e
2+α( e Ω). Let b Ω = Φ( e Ω) = {z : |z
i| < d, i = 1, 2, 0 < z
3< d} and S = Φ( e b S). Define
(2.4) u(z) = u b
′(y)|
y=Φ−1(z), q(z) = q b
′(y)|
y=Φ−1(z).
Introduce b ∇
k= ξ
lxkz
iξl∇
zi|
ξ=χ−1(z), where χ(ξ) = Φ(ψ
0(ξ)) and y = ψ
0(ξ) is described by (2.1). Introduce also the notation
e
u(ξ) = u(ξ)ζ(ξ), q e
σ(ξ) = q
σ(ξ)ζ(ξ), ξ ∈ e Ω, e Ω ∩ S = ∅, (2.5)
e
u(z) = b u(z)b ζ(z), q e
σ(z) = b q
σ(z)b ζ(z), z ∈ b Ω = Φ( e Ω), e Ω ∩ S 6= ∅, (2.6)
where b ζ(z) = ζ(ξ)|
ξ=χ−1(z).
Under the above notation problem (1.14) has the following form in an interior subdomain:
(2.7)
ηe u
it− ∇
jT
ij(e u, e q
σ) = ηeg
i− ∇
ujB
uij(u, ζ) − T
uij(u, q
σ)∇
ujζ
− (∇
jT
ij(e u, e q
σ) − ∇
ujT
uij(e u, e q
σ)) ≡ ηeg
i+ k
1i, i = 1, 2, 3, e
q
σt+ p
1∇ · e u = p
1u · ∇
uζ + (p
1− Ψ (η))∇
u· uζ + p
1(∇ · e u − ∇
u· e u) ≡ k
2,
where Ψ (η) = p
η(η)η, p
1= Ψ (̺
e), and in a boundary subdomain:
(2.8) b
η e u
it− ∇
jT
ij(e u, e q
σ) = b η eg
i− b ∇
jB b
ij(b u, b ζ) − b T
ij(b u, b q
σ) b ∇
jζ b
− (∇
jT
ij(e u, e q
σ) − b ∇
jT b
ij(e u, e q
σ)) ≡ b η eg
i+ k
3i, i = 1, 2, 3, e
q
σt+ p
1∇ · e u = p
1b u · b ∇b ζ + (p
1− Ψ (b η)) b ∇ · e ub ζ + p
1(∇ · e u − b ∇ · e u) ≡ k
4, b
n · b D (e u) · b τ
α= b n · b B (b u, b ζ) · b τ
α− γb u · b τ
α≡ k
5α, α = 1, 2, b
n · e u = 0,
where the summation convention is assumed,
(2.9) B
uij(u, ζ) = µ(u
i∇
ujζ + u
j∇
uiζ) + (ν − µ)δ
iju · ∇
uζ,
and b T , b B indicate that the operator ∇
uis replaced by b ∇. Finally the dot · denotes the scalar product in R
3.
Let ψ(y) ≡ y
3− e F (y) = 0. Then b n = ∇ψ/|∇ψ|, τ
1= (0, 1, 0) × b n,
τ
2= (1, 0, 0) × b n. Moreover, in the next considerations we denote z
1, z
2by τ , and z
3by n. From [11] we recall the Korn inequality. For vectors
u, v ∈ H
1(Ω) we introduce (2.10) E(u, v) =
\
Ω
(∂
xiu
j+ ∂
xju
i)(∂
xiv
j+ ∂
xjv
i) dx.
The vectors which satisfy E(u, u) = 0 have the form
(2.11) u = A + B × x,
where A, B are arbitrary constant vectors.
We define e H(Ω) = {u : E(u, u) < ∞, u · n = 0 on S}. If Ω is a domain obtained by rotation about the vector B we set H(Ω) = {u ∈ e H(Ω) :
T
Ω
u · B × x dx = 0}. Otherwise we set H(Ω) = e H(Ω).
Lemma 2.1 (see [11]). Let S ∈ C
2. Then for u ∈ H(Ω), (2.12) kuk
21,Ω≤ c
0E(u, u).
Finally, we introduce the notation. By H
k+α,k/2+α/2(Ω
T), k ∈ N ∪ {0}, α ∈ (1/2, 1), we denote the Hilbert space with the norm
kuk
2Hk+α,k/2+α/2(ΩT)= X
|β|+2i≤k
k∂
xβ∂
tiuk
2L2(ΩT)
+ X
|β|=k T
\
0
\
Ω
\
Ω
|∂
xβu(x, t) − ∂
xβ′u(x
′, t)|
2|x − x
′|
3+2αdx dx
′dt
+
\
Ω T
\
0 T
\
0
|∂
t[k/2]u(x, t) − ∂
t[k/2]′u(x, t
′)|
2|t − t
′|
1+α+k−2[k/2]dx dt dt
′. Similarly we introduce the spaces H
k+α(Ω), H
k+α,k/2+α/2(S
T). Moreover, we use the notation
kuk
Hk+α,k/2+α/2(ΩT)= kuk
k+α,ΩT,
kuk
Hk+α(Ω)= kuk
k+α,Ω, kuk
Hk+α(0,T )= kuk
k+α,(0,T ), kuk
Lp(Ω)= |u|
p,Ω, p ∈ [1, ∞],
kuk
Hs(ΩT)= kuk
s,ΩT, s ∈ N ∪ {0}.
The spaces H
k+α,k/2+α/2(Ω
T) are Sobolev–Slobodetski˘ı spaces and are de- noted also by W
k+α,k/2+α/22
(Ω
T). They are particular cases of Besov spaces.
First we recall some properties of isotropic Besov spaces which are fre- quently used in this paper; next we define anisotropic Besov spaces and formulate some imbedding theorems which we need.
Let us introduce the differences
∆
i(h)u(x) = u(x + he
i) − u(x),
where x ∈ R
nand e
i, i = 1, . . . , n, are the standard unit vectors. Then we define inductively the m-difference
∆
mi(h)u(x) = ∆
i(h)(∆
m−1i(h)u(x)) = X
m j=0(−1)
m−jc
jmu(x + jhe
i),
where c
jm=
mj=
j!(m−j)!m!. Moreover, we introduce the difference
∆(y)f (x) = f (x + y) − f (x), x, y ∈ R
n, and inductively
∆
m(y)f (x) = ∆(y)(∆
m−1(y)f (x)).
Since
∆(x − y)f (y) = f (x) − f (y) we have
∆
m(x − y)f (y) = X
ni=1
∆
m((x − y) · e
i)f (y) = X
n i=1∆
mi(h)f (y), where the last equality holds for (x − y) · e
i= h.
We define the isotropic Besov spaces by introducing the norm (see [2, Sect. 18])
(2.13) kuk
Bpl(Rn)= kuk
Lp(Rn)+ X
n i=1 h\00
dh
\
Rn
dx |∆
mi(h)∂
xkiu|
ph
1+(l−k)p 1/p, where m > l − k, m, k ∈ N ∪ {0}, l ∈ R
+and l 6∈ Z.
It was shown in [4] that the Besov spaces defined by (2.13) all coincide and have equivalent norms for all m, k satisfying m > l − k.
Next we define the L
p-scale of Sobolev–Slobodetski˘ı spaces by introduc- ing the norm
(2.14) kuk
Wpl(Rn)= kuk
Lp(Rn)+ X
n i=1 h\00
dh
\
Rn
dx |∆
i(h)∂
x[l]iu|
ph
1+p(l−[l]) 1/p, where l 6∈ Z, [l] is the integral part of l. We frequently write l = k + α, k ∈ N ∪ {0}, α ∈ (0, 1), so k = [l] and α = l − [l].
By the Golovkin theorem [4] the norms of the spaces B
pl(R
n) and W
pl(R
n) are equivalent.
We also define the norms (2.15) kuk
Belp(Rn)
= kuk
Lp(Rn)+
\Rn
dx
\
Rn
dy |∆
m(x − y)∂
yku(y)|
p|x − y|
n+p(l−k) 1/pfor any m > l − k (here ∂
yku = P
|α|=k
D
yαu) and (2.16) kuk
Wflp(Rn)
= kuk
Lp(Rn)+
\Rn
dx
\
Rn
dy |∆(x − y)∂
y[l]u(y)|
p|x − y|
n+p(l−[l]) 1/p.
Using Lemma 7.44 of [1] we can show that the spaces e B
pl(R
n) and B
pl(R
n) (with h
0= ∞) and f W
pl(R
n) and W
pl(R
n) (with h
0= ∞) all coincide and have equivalent norms.
Therefore all norms (2.13)–(2.16) are equivalent. Now we recall some imbedding theorems for Besov spaces (see [2, Sect. 18]):
(2.17) D
σB
pl(R
n) ⊂ B
̺q(R
n) for n/p − n/q + σ + ̺ ≤ l.
For κ = 1 −
1l np−
nq+ σ + ̺) > 0 we have the interpolation inequality (2.18) k∂
σxuk
B̺q(Rn)≤ ε
1−κkuk
Bpl(Rn)+ cε
−κkuk
Lp0(Rn),
with p
0≥ 1, ε ∈ (0, 1). In the above notation B
pl(R
n) with l ∈ Z
+is a Sobolev space and L
p(R
n) = W
p0(R
n).
All the above remarks can be applied to spaces of functions defined on a bounded domain Ω ⊂ R
n. Moreover by using a partition of unity we can define spaces of traces on the boundary of Ω and formulate the corresponding trace theorems.
Now we introduce some anisotropic Sobolev–Slobodetski˘ı and Besov spaces which are necessary for our considerations. First we define
kuk
Lp1,p2(ΩT)=
T\0
dt
\Ω
dx |u(x, t)|
p1p2/p11/p2and
kuk
Lp1,p2(ΩT)=
\Ω
dx
T\0
dt |u(x, t)|
p1p2/p11/p2, where p
i∈ [1, ∞], i = 1, 2.
We need imbedding theorems between the above spaces and W
2l,l/2(Ω
T), l ∈ R
+, Ω ⊂ R
n. We apply the results from [2, Sect. 18]. Since W
2l,l/2(Ω
T) is isotropic with respect to the power of integration we have, for 2 ≤ p
1, p
2< ∞,
(2.19) ∂
xα1∂
tα2W
2l,l/2(Ω
T) ⊂ L
p1,p2(Ω
T) if n 2 − n
p
1+ 2 2 − 2
p
2+ α
1+ 2α
2≤ l, (2.20) ∂
xα1∂
tα2W
2l,l/2(Ω
T) ⊂ L
p1,p2(Ω
T) if n
2 − n p
2+ 2 2 − 2
p
1+ α
1+ 2α
2≤ l,
(2.21) ∂
xα1∂
tα2W
2l,l/2(Ω
T) ⊂ L
p(0, T ; B
qσ(Ω)) if n
2 − n q + 2
2 − 2
p + α
1+ 2α
2+ σ ≤ l, and finally
(2.22) ∂
xα1∂
αt2W
2l,l/2(Ω
T) ⊂ L
p(Ω; B
qσ(0, T )) if n
2 − n p + 2
2 − 2
q + α
1+ 2α
2+ 2σ ≤ l, 2 ≤ p, q < ∞.
Moreover, the corresponding interpolation inequalities hold.
3. Inequality for global existence. First we obtain an energy type inequality:
Lemma 3.1. Assume that (v, ̺
σ) is the solution to problem (1.11) deter- mined by Theorem 1.2. Let the assumptions of Lemma 2.1 hold. Then (3.1) d
dt
\
Ω
1
2 ̺v
2+ p
2σ2p
̺̺
dx + c
0kvk
21,Ω+ γ X
2 α=1\
S
|v · τ
α|
2dS
≤ ψ
1(|̺|
∞,ΩT, |1/̺|
∞,ΩT)kp
σk
41,Ω+
\
Ω
̺f · v dx, where ψ
1is an increasing positive function.
P r o o f. Multiplying (1.11)
1by v, integrating over Ω, using the boundary conditions and the continuity equation we obtain
(3.2) d dt
1 2
\
Ω
̺v
2dx +
\
Ω
D (v) · ∇v dx
−
\
Ω
p
σdiv v dx + γ X
2 α=1\
S
|v · τ
α|
2dS =
\
Ω
̺f · v dx.
In view of the Korn inequality (see Lemma 2.1), using (1.12) in the third term on the l.h.s. of (3.2) yields
(3.3) d dt
1 2
\
Ω
̺v
2dx + c
0kvk
21,Ω+
\
Ω
p
σp
̺̺ (p
σt+ v · ∇p
σ) dx + γ
X
2 α=1\
S
|v · τ
α|
2dS ≤
\
Ω
̺f · v dx,
where c
0is the constant from the Korn inequality.
Continuing calculations in the third term on the l.h.s. of (3.3) we get
\
Ω
1
2p
̺̺ (p
2σ,t+ v · ∇p
2σ) dx =
\
Ω
̺
p
2σ2p
̺̺
2t
+ ̺v · ∇
p
2σ2p
̺̺
2dx
−
\
Ω
1 2 ̺p
2σ1 p
̺̺
2p
(p
σt+ v · ∇p
σ) dx
= d dt
\
Ω
p
2σ2p
̺̺ dx + N
1, where
N
1= 1 2
\
Ω
p
̺̺
21 p
̺̺
2p
p
2σdiv v dx, so
|N
1| ≤ εk div vk
20,Ω+ ϕ
1(1/ε, |̺|
∞,ΩT, |1/̺|
∞,ΩT)kp
σk
41,Ω,
where ϕ
1is an increasing positive function. Using the above considerations in (3.3) we obtain (3.1). This concludes the proof.
Now we show
Lemma 3.2. For the local solution determined by Theorems 1.1, 1.2 we have
ϕ(t, Ω) + kuk
22+α,Ωt+ kq
σk
21+α,Ωt≤ c(kuk
20,Ωt+ kq
σk
20,Ωt) (3.4)
+ ckgk
2α,Ωt+ cX
12+ ϕ(0, Ω), where t ≤ T , T is the time of local existence, and ϕ(t, Ω) and X
1are defined by (3.62) and (3.64), respectively.
P r o o f. First we obtain the inequality for interior subdomains. Assume that ξ
0∈ e Ω, where e Ω is an interior subdomain, and ζ ∈ C
0∞(R
3) is the corresponding function from the partition of unity such that ζ(ξ) = 1 for ξ ∈ B
λ(ξ
0) = {ξ ∈ R
3: |ξ − ξ
0| < λ} and ζ(ξ) = 0 for ξ ∈ R
3\ B
2λ(ξ
0).
Denote by ∆
s(z)f (x) the sth finite difference of f such that
∆
s(z)f (x) = X
s k=0c
ks(−1)
s−kf (x + kz), where c
ks=
sk=
k!(s−k)!s!.
In this proof, functions ϕ
j, ψ
j, j ∈ N ∪ {0}, are increasing continuous
positive functions of their arguments. Applying ∆
s(z) to (2.7)
1, multiplying
the result by ∆
s(z)e u
i, integrating over B
2λ(ξ
0) and integrating with respect
to z over R
3with the weight 1/|z|
3+2(1+α)we obtain
(3.5)
\
R3
dz
\
B2λ(ξ0)
dξ η∆
s(z)e u
it∆
s(z)e u
i|z|
3+2(1+α)−
\
R3
dz
\
B2λ(ξ0)
dξ ∇
ξj(∆
s(z)T
ij(e u, e q
σ))∆
s(z)e u
i|z|
3+2(1+α)= −
\
R3
dz
\
B2λ(ξ0)
dξ (∆
s(z)(ηe u
it) − η∆
s(z)e u
it)∆
s(z)e u
i|z|
3+2(1+α)+
\
R3
dz
\
B2λ(ξ0)
dξ ∆
s(z)(ηeg
i+ k
1i)∆
s(z)e u
i|z|
3+2(1+α)≡ I
1+ I
2. Using the continuity equation (1.15)
1in the first term and integrating by parts in the second with the use of the boundary condition e u|
∂B2λ(ξ0)= 0 we obtain
(3.6) 1 2
d dt
\
R3
dz
\
B2λ(ξ0)
dξ η|∆
s(z)e u|
2|z|
3+2(1+α)+
\
R3
dz
\
B2λ(ξ0)
dξ D (∆
s(z)e u) · ∇∆
s(z)e u
|z|
3+2(1+α)−
\
R3
dz
\
B2λ(ξ0)
dξ ∆
s(z)e q
σdiv(∆
s(z)e u)
|z|
3+2(1+α)= I
1+ I
2− 1 2
\
R3
dz
\
B2λ(ξ0)
dξ η div
uu|∆
s(z)e u|
2|z|
3+2(1+α)≡ X
3 i=1I
i.
To examine the last term on the l.h.s. of (3.6) we use (2.7)
2. Applying
∆
s(z) to (2.7)
2, multiplying the result by ∆
s(z)e q
σ, integrating over B
2λ(ξ
0) and integrating with respect to z with the weight 1/|z|
3+2(1+α)we obtain (3.7) 1
2 d dt
\
R3
dz
\
B2λ(ξ0)
dξ |∆
s(z)e q
σ|
2|z|
3+2(1+α)+ p
1\
R3
dz
\
B2λ(ξ0)
dξ ∆
s(z)e q
σdiv ∆
s(z)e u
|z|
3+2(1+α)=
\
R3
dz
\
B2λ(ξ0)
dξ ∆
s(z)k
2∆
s(z)e q
σ|z|
3+2(1+α)≡ I
4.
Multiplying (3.7) by 1/p
1, adding to (3.6) and using the Korn inequality we get
(3.8) 1 2
d dt
\
R3
dz
\
B2λ(ξ0)
dξ
η|∆
s(z)e u|
2|z|
3+2(1+α)+ 1
p
1· |∆
s(z)e q
σ|
2|z|
3+2(1+α)+ 1 c
0\
R3
dz k∆
s(z)e uk
21,B2λ(ξ0)
|z|
3+2(1+α)≤ X
3 i=1I
i+ 1 p
1I
4. Now we estimate the terms on the r.h.s. of (3.8). Since α ∈ (1/2, 1) and s must be such that s > 1 + α (see [4]) we assume that s = 2. First we consider
I
2=
\
R3
dz
\
B2λ(ξ0)
dξ ∆
2(z)(ηeg
i+ k
1i) · ∆
2(z)e u
i|z|
3+2(1+α)= −
\
R3
dz
\
B2λ(ξ0)
dξ ∆(z)(ηeg
i+ k
1i) · ∆
3(z)e u
i|z|
3+2(1+α)≤ ε
1\
R3
dz
\
B2λ(ξ0)
dξ |∆
3(z)e u|
2|z|
3+2(2+α)+ c(ε
1)
\
R3
dz
\
B2λ(ξ0)
dξ |∆(z)(ηeg + k
1)|
2|z|
3+2α≡ I
21+ I
22, where
I
22≤ c
\
R3
dz
\
B2λ(ξ0)
dξ 1
|z|
3+2α(|∆(z)(ηeg)|
2+ |∆(z)∇
uB
u(u, ζ)|
2+ |∆(z)T
u(u, q
σ)∇
uζ|
2+ |∆(z)(∇T(e u, e q
σ) − ∇
uT
u(e u, e q
σ))|
2) ≡ X
4 i=1I
5i. Next we estimate
I
51≤ c
\
R3
dz
\
B2λ(ξ0)
dξ |∆(z)ηeg|
2+ |η∆(z)eg|
2|z|
3+2α≤ c
\R3
dz
\
B2λ(ξ0)
dξ |∆(z)η|
2p|z|
3+2p(α+ε/2) 1/p \R3
dz
\
B2λ(ξ0)
dξ |eg(ξ, t)|
2p′|z|
3−p′ε 1/p′+ c|η|
∞,Ω\
R3
dz
\
B2λ(ξ0)
dξ |∆(z)eg(ξ, t)|
2|z|
3+2α,
for all p, p
′with 1/p+1/p
′= 1 and ε > 0. Using the imbeddings B
21+α(Ω) ⊂
B
2pα+ε/2(Ω), Ω ⊂ R
3, valid if 3/2 − 3/(2p) + α + ε/2 < 1 + α, and B
2α(Ω) ⊂
L
2p′(Ω), Ω ⊂ R
3, valid if 3/2 − 3/(2p
′) < α, which both hold for suitable p,
ε because 3/2 + ε/2 < 1 + α can be satisfied for α > 1/2, we obtain I
51≤ ckηk
21+α,Ωkegk
2α,Ω,
where we used the fact that B
21+α(Ω) ⊂ L
∞(Ω) for α > 1/2.
Introducing the quantity α(t) =
Tt
0
u
ξ(ξ, τ ) dτ we see that ξ
x− I = αψ
0(α), ξ
x2− I = αψ
1(α),
ξ
x∇
ξξ
x= ∇
ξαψ
2(α) + α∇
ξαψ
3(α), where ξ
x, I are matrices, I is the unit matrix,
|α(t)|
2≤ T
t
\
0
|u
ξ|
2dτ ≤ cT
t
\
0
kuk
22+α,Ωdτ, and
a ≡ cT
1/2T\0
kuk
22+α,Ωdτ
1/2so
|ξ
x− I|
2, |ξ
x2− I|
2≤ a
2ψ
4(a), (3.8
′)
|ξ
x∇
ξξ
x|
2≤ ψ
5(a)
T
\
0
|u
ξξ|
2dτ.
Using the H¨older inequality and the imbeddings B
2α(Ω) ⊂ L
2p(Ω), B
21+α(Ω) ⊂ B
α2p′(Ω), Ω ⊂ R
3, holding for 3/2 − 3/(2p) ≤ α, 3/2 − 3/(2p
′) + α ≤ 1 + α, 1/p + 1/p
′= 1, we have
I
52= c
\
R3
dz
\
B2λ(ξ0)
|∆(z)∇
uB
u(uϕ)|
2|z|
3+2αdξ
≤ c
\
R3
dz
\
B2λ(ξ0)
dξ
× 1
|z|
3+2α[|∆(z)∇u|
2+ ξ
x4|∇ζ|
2+ |∇u|
2(|∆(z)ξ
x2|
2|∇ζ|
2+ ξ
x2|∆(z)∇ζ|
2) + |∆(z)u|
2(|ξ
x· ∇ξ
x|
2|∇ζ|
2+ ξ
x2|∇
2ζ|
2) + |u|
2(|∆(z)(ξ
x· ∇ξ
x)|
2|∇ζ|
2+ |ξ
x· ∇ξ
x|
2|∆(z)∇ζ|
2+ |∆(z)ξ
x2|
2|∇
2ζ|
2+ ξ
4x|∆(z)∇
2ζ|
2)]
≤ ψ
6(a) h
kuk
21+α,B2λ(ξ0)+ (ku
ξk
2Bα2p′(B2λ(ξ0))
+ kuk
2Bα2p′(B2λ(ξ0))
)
t
\
0
ku
ξξk
2L2p(B2λ(ξ0))dτ i + ψ
7(a)(|∇u|
2∞,B2λ(ξ0)
+ |u|
2∞,B2λ(ξ0)
)
× h
\t0
kuk
22+α,B2λ(ξ0)dτ +
t
\
0
ku
ξk
2Bα2p′(B2λ(ξ0))
dτ
t
\
0
ku
ξξk
2L2p(B2λ(ξ0))dτ i
≤ cϕ
0(a)kuk
21+α,Ω+ ϕ
1(a)kuk
22+α,Ωt
\
0
kuk
22+α,Ωdτ.
Similarly, we have
I
53≤ ϕ
2(a)(kuk
21+α,Ω+ kq
σk
2α,Ω).
Since qualitatively
∇T(e u, e q
σ) − ∇
uT
u(e u, e q
σ) = ϕ
3(α)
t
\
0
u
ξξ(τ ) dτ e u
ξ+ αϕ
4(α)(e u
ξξ+ e q
σξ), where ϕ
i(0) 6= 0, i = 3, 4, we obtain
I
54≤ ϕ
5(a)
t
\
0
kuk
22+α,Ωdτ (ke uk
22+α,Ω+ ke q
σk
21+α,Ω).
Summarizing the above considerations we have
|I
2| ≤ ε
1ke uk
22+α, eΩ+ ϕ
6(1/ε
1, a)(kuk
21+α, eΩ+ kq
σk
2α, eΩ) (3.9)
+ ϕ
7(1/ε
1, a)
t
\
0
kuk
22+α,Ωdτ (ke uk
22+α, eΩ+ ke q
σk
21+α, eΩ), where ε
1∈ (0, 1).
Now we calculate I
1= −
\
R3
dz
\
B2λ(ξ0)
dξ (∆
2(z)ηe u
t+ 2∆(z)η∆(z)e u
t)∆
2(z)e u
|z|
3+2(1+α)= −
\
R3
dz
\
B2λ(ξ0)
dξ (∆
2(z)η
σe u
t+ 2∆(z)η
σ∆(z)e u
t)∆
2(z)e u
|z|
3+2(1+α).
We have
|I
1| ≤ c
\R3
dz
\
B2λ(ξ0)
dξ |∆
2(z)η
σ|
2|z|
3+2(1+α) 1/2 \R3
dz
\
B2λ(ξ0)
dξ |e u
t|
p|z|
3−εp 1/p×
\R3
dz
\
B2λ(ξ0)
dξ |∆
2(z)e u|
p′|z|
3+p′(1+α)+εp′ 1/p′+ c
\R3
dz
\
B2λ(ξ0)
dξ |∆(z)η
σ|
p1|z|
3+(p1/2)(1+α) 1/p1×
\R3
dz
\
B2λ(ξ0)
dξ |∆(z)e u
t|
p2|z|
3+(p2/2)(1+α) 1/p2×
\R3
dz
\
B2λ(ξ0)
dξ |∆
2(z)e u|
p3|z|
3+p3(1+α) 1/p3≡ I
11+ I
12,
for all p, p
iwith 1/p + 1/p
′= 1/2 and 1/p
1+ 1/p
2+ 1/p
3= 1.
Using the imbeddings B
2α(Ω) ⊂ L
p(Ω), B
22+α(Ω) ⊂ B
p1+α+ε′(Ω), valid if 3/2 − 3/p ≤ α, 3/2 − 3/p
′+ 1 + α + ε ≤ 2 + α, which both hold for suitable p, ε because 3/2 + ε ≤ 1 + α can be satisfied for α > 1/2, we obtain
I
11≤ ckη
σk
1+α,Ωke u
tk
α, eΩke uk
2+α, eΩ≤ ε
2ke uk
22+α, eΩ+ c(ε
2)kη
σk
21+α,Ωke u
tk
2α, eΩ.
Using the imbeddings B
21+α(Ω) ⊂ B
p(1+α)/21(Ω), B
2α(Ω) ⊂ B
p(1+α)/22(Ω), B
22+α(Ω) ⊂ B
1+αp3(Ω) valid for 3/2 − 3/p
1+ (1 + α)/2 ≤ 1 + α, 3/2 − 3/p
2+ (1 + α)/2 ≤ α, 3/2 − 3/p
3+ 1 + α ≤ 2 + α, which all hold for suitable p
ibecause 3/2 ≤ 1 + α, we get the same bound as for I
11, I
12≤ ε
2ke uk
22+α, eΩ+ c(ε
2)kη
σk
21+α,Ωke u
tk
2α, eΩ. Next we have
|I
3| ≤ ε
3ke uk
22+α, eΩ+ c(ε
3)kuk
22+α,Ωke uk
21+α, eΩ. Finally, we have
|I
4| ≤ ε
4\
R3
dz
\
B2λ(ξ0)
dξ |∆
2(z)e q
σ|
2|z|
3+2(1+α)+ ϕ
8(1/ε
4, a) h
kuk
21+α, eΩ+ kuk
22+α, eΩt
\
0
kuk
22+α,Ωdτ + kq
σk
21+α, eΩkuk
22+α, eΩ+ kq
σk
21+α, eΩkuk
22+α, eΩt
\
0
kuk
22+α,Ωdτ + kq
σk
21+α, eΩt
\
0
kuk
22+α,Ωdτ i .
Summarizing the above considerations we obtain (3.10) d
dt
\
R3
dz
\
B2λ(ξ0)
dξ
η
2 · |∆
2(z)e u|
2|z|
3+2(1+α)+ 1
2p
1· |∆
2(z)e q
σ|
2|z|
3+2(1+α)+ 1 c
0\
R3
dz k∆
2(z)e uk
21,B2λ(ξ0)
|z|
3+2(1+α)≤ ε
5(ke uk
22+α, eΩ+ ke q
σk
21+α, eΩ)
+ ϕ
9(1/ε
5, a)(kuk
21+α, eΩ+ kq
σk
2α, eΩ)
+ ϕ
10(1/ε
5, a)
t
\
0
kuk
22+α,Ωdτ (kuk
22+α, eΩ+kq
σk
21+α, eΩ+kq
σk
21+α, eΩkuk
22+α, eΩ) + ϕ
11(1/ε
5, a)(kq
σk
21+α, eΩke u
tk
2α, eΩ+ kuk
22+α, eΩke uk
21+α, eΩ+ kq
σk
21+α, eΩkuk
22+α, eΩ+ kegk
2α, eΩ).
Now we want to obtain an energy type estimate for the local solutions satisfying (2.7). Multiplying (2.7)
1by e u
i, summing over i, integrating over Ω and using the boundary condition e
(3.11) e u|
∂ eΩ= 0,
and employing the continuity equation, we obtain (3.12)
\
e Ω
η 1 2
d
dt e u
2dξ +
\
e Ω
D (e u)∇e u dξ −
\
e Ω
e
q
σ∇ · e u dξ =
\
e Ω
(ηeg + k
1)e u dξ.
Using the continuity equation (1.15)
1and the Korn inequality we get (3.13) d
dt
\
Ωe
1
2 ηe u
2dξ + 1
c
0ke uk
21, eΩ−
\
Ωe
e
q
σ∇ · e u dξ
≤ − 1 2
\
Ωe
η div
uue u
2dξ +
\
Ωe
(ηeg + k
1)e u dξ.
Multiplying (2.7)
2by e q
σand integrating over e Ω yields
(3.14) 1
2 d dt
\
Ωe
e
q
σ2dξ + p
1\
Ωe
e
q
σ∇ · e u dξ =
\
Ωe
k
2e q
σdξ.
From (3.13) and (3.14) we obtain (3.15) d
dt
\
Ωe
1 2
ηe u
2+ 1 p
1q e
σ2dξ + 1
c
0ke uk
21, eΩ≤ 1 p
1\
Ωe
k
2q e
σdξ − 1 2
\
Ωe
η div
uue u
2dξ +
\
Ωe
(ηeg + k
1)e u dξ
≡ J
1+ J
2+ J
3, where
|J
1| ≤ εke q
σk
20, eΩ+ ϕ
12(1/ε, a)
kuk
20, eΩ+ kq
σk
21, eΩkuk
22, eΩ+
t
\
0
kuk
22+α,Ωdτ ke uk
21, eΩ,
|J
2| ≤ εke uk
21, eΩ+ ϕ
13(1/ε, a)kuk
21, eΩke uk
21, eΩ,
|J
3| ≤ ε(ke uk
21, eΩ+ kq
σk
20, eΩ)
+ ϕ
14(1/ε, a) h
kuk
21, eΩ+
t
\
0
kuk
22, eΩdτ (ke uk
22, eΩ+ ke q
σk
21, eΩ) i
+ ckegk
20, eΩ. Using the above estimates in (3.15) gives
(3.16) d dt
\
Ωe
ηe u
2+ 1 p
1q e
σ2dξ + 1
c
0ke uk
21, eΩ≤ εkq
σk
20, eΩ+ ϕ
15(1/ε, a) h
kegk
20, eΩ+ kuk
21, eΩ+ kuk
21, eΩke uk
21, eΩ+ kq
σk
21, eΩkuk
22, eΩ+
t
\
0
kuk
22+α,Ωdt (kuk
22, eΩ+ kq
σk
21, eΩ) i . From (3.10) and (3.16) after using the equivalence of norms for fractional spaces (see [4]) we obtain
(3.17) d dt
\
e Ω
ηe u
2+ 1 p
1q e
σ2dξ
+ d dt
\
R3
dz
\
B2λ(ξ0)
dξ
η |∆
2(z)e u|
2|z|
3+2(1+α)+ 1 p
1· |∆
2(z)e q
σ|
2|z|
3+2(1+α)+ ke uk
22+α, eΩ≤ εke q
σk
21+α, eΩ+ c(kuk
21+α, eΩ+ kq
σk
2α, eΩ)
+ c[kegk
2α, eΩ+ kq
σk
21+α, eΩke u
tk
2α, eΩ+ kuk
22+α, eΩke uk
21+α, eΩ+ kq
σk
21+α, eΩkuk
22+α, eΩ]
+ c
t
\
0
kuk
22+α,Ωdτ [kuk
22+α, eΩ+ kq
σk
21+α, eΩ+ kq
σk
21+α, eΩkuk
22+α, eΩ].
To examine the second term on the r.h.s. of (3.17) we use the interpolation inequality
kuk
21+α, eΩ+ kq
σk
2α, eΩ≤ ε(kuk
22+α, eΩ+ kq
σk
21+α, eΩ) (3.18)
+ c(ε)(kuk
20, eΩ+ kq
σk
20, eΩ).
From (2.7)
1we have
ke q
σk
21+α, eΩ≤ c(ke u
tk
2α, eΩ+ ke uk
22+α, eΩ+ kegk
2α, eΩ) (3.19)
+ c
t
\
0
kuk
22+α,Ωdτ (ke uk
22+α, eΩ+ ke q
σk
21+α, eΩ)
+ ε(kuk
22+α, eΩ+ kq
σk
21+α, eΩ) + c(ε)(kuk
20, eΩ+ kq
σk
20, eΩ).
From (3.17)–(3.19) we obtain (3.20) d
dt
\
Ωe
ηe u
2+ 1 p
1q e
σ2dξ
+ d dt
\
R3
dz
\
B2λ(ξ0)
dξ
η|∆
2(z)e u|
2|z|
3+2(1+α)+ 1 p
1· |∆
2(z)e q
σ|
2|z|
3+2(1+α)+ ke uk
22+α, eΩ+ ke q
σk
21+α, eΩ≤ εke u
tk
2α, eΩ+ ckegk
2α, eΩ+ ε(kuk
22+α, eΩ+ kq
σk
21+α, eΩ)c(ε)(kuk
20, eΩ+ kq
σk
20, eΩ) + c[kq
σk
21+α, eΩke u
tk
2α, eΩ+ kuk
22+α, eΩke uk
21+α, eΩ+ kq
σk
21+α, eΩkuk
22+α, eΩ] + c
t
\
0
kuk
22+α,Ωdτ [kuk
22+α, eΩ+ kq
σk
21+α, eΩ+ kq
σk
21+α, eΩkuk
22+α, eΩ].
To examine the norm ke u
tk
α, eΩfrom the r.h.s. of (3.20) we use the inter- polation inequality (see [12])
(3.21)
t
\
0
kω
tk
2α, eΩdt ≤ ε
t
\
0
kwk
22+α, eΩdt + c(ε)
\
Ωe
kw
tk
2α/2,(0,t)dξ.
To estimate the last term on the r.h.s. of (3.21) we consider the time differ- ences
∆
kt(h)f (ξ, t) = X
k j=0c
jk(−1)
k−jf (ξ, t + jh), k > α.
Applying ∆
kt(h) to (2.7)
1we obtain
(3.22) η∆
kte u
it− ∇
jT
ij(∆
ktu, ∆ e
ktq e
σ) = η∆
ktu e
it− ∆
kt(ηe u
it) + ∆
kt(ηeg
i+ k
1i).
From (3.22) we have (3.23)
h\0
0
dh h
1+αT −kh\0
0
ψ(t) dt
\
Ωe
η|∆
kte u
t|
2dξ
−
h\0
0
dh h
1+αT −kh\0
0
ψ(t) dt
\
Ωe
∇T(∆
kte u, ∆
ktq e
σ)∆
ktu e
tdξ
=
h0
\
0
dh h
1+αT −kh0
\
0
ψ(t) dt
\
e Ω