Morrey space regularity for weak solutions of Stokes systemswith VMO coefficients

Josef Danecek, Oldrich John and Jana Stara ∗

27.11.2009

1 Introduction.

In this paper we prove that weak solutions (u, p) ∈ W 1,20 (Ω,Rd)× L2(Ω,R) of Stokes system

−divAEu+∇ p = −divf (1)

div u = 0

on a domain Ω have the symmetric gradient Eu and pressure p in Morrey’s space L2,µ(Ω) if thecoefficients are essentially bounded, have vanishing mean oscillations and the right hand side lies inthe same Morrey’s space.Stationary and nonstationary Stokes systems, their solvability and properties of solutions form thefundamental part of fluid dynamics. We consider here coefficients that need not be necessarily contin-uous in the classical sense and satisfy only a kind of integral continuity condition. The correspondingfunction space VMO(Ω) was introduced by D. Sarason (see [14]) as a proper subspace of frequentlyused John-Nirenberg space BMO(Ω) of functions with bounded mean oscillations. Several authors([15], [3], [4]) proved that this type of coefficients is sufficient for higher regularity of solutions to ellip-tic systems. In our proof we substantially use the method of A-harmonic approximations introducedby Duzaar and Grotowski in [5] .

2 Preliminaries and Notations.

In the whole paper Ω is an open and bounded domain with C1-boundary in Rd,(d ≥ 2). By BR(x) wedenote an open ball in Rd centered in x with radius R. For x = [x1, . . . , xd−1, 0] we denote B+

R(x) =y = [y1, ..., yd−1, yd] ∈ Rd; |x− y| < R; yd > 0 and ΓR(x) = y = [y1, ..., yd−1, 0] ∈ Rd; |x− y| < R.Let ΩR(x) = Ω ∩ BR(x). For a function f integrable on ΩR(x) having positive measure we denote(f)ΩR(x) = 1

|ΩR(x)|

∫ΩR(x)

f(y) dy its integral mean value.

1991 Mathematics Subject Classifications: 35J60Key words and phrases: Stokes systems, linear elliptic systems, regularity, Morrey spaces, VMO spaces

∗J.Danecek was supported by the research project MSM0021630511, Oldrich John and Jana Stara were supportedby the research project MSM0021620839 and Jana Stara by GACR 201/09/0917.

1

By C∞0 (Ω,Rd) we denote standard space of infinitely differentiable functions with compact support in

Ω and by W 1,2(BR(x),Rd) and W 1,20 (BR(x),Rd) we denote Sobolev spaces. Moreover, for a function

space X(Ω,Rd) the subscript ”div” denotes

Xdiv(Ω,Rd) = u ∈ X(Ω,Rd); div u = 0 on Ω.

Definition(Morrey space)Let q ≥ 1 and 0 ≤ µ < n. By Lq,µ(Ω,Rd) we denote the linear space formed by all the functionsu ∈ Lq(Ω,Rd) for which

‖u‖Lq,µ(Ω) = supx∈Ω, 0<ρ≤diam(Ω)

ρ−µ

∫Ωρ(x)

|u(y)|q dy

1/q

< +∞.

Lq,µ(Ω,Rd) equipped with the above norm is a Banach space.

Definition(VMO space)For a function w ∈ L1(Ω,Rd) and r > 0, x ∈ Ω we define

η(x, r) ≡ sup0<ρ≤r

1

|Ωρ(x)|

∫Ωρ(x)

∣∣w(y)− (w)Ωρ(x)

∣∣ dy.By VMO = VMO(Ω,Rd) we denote the space of all functions w ∈ L1(Ω,Rd) such that

η(r) ≡ supx∈Ω

η(x, r) < +∞ for all 0 < r ≤ diam(Ω)

andlim

r→0+η(r) = 0.

For a function u ∈ W 1,2(Ω,Rd) denote by∇u = ( ∂ui

∂xj)i,j=1,...,d its gradient and by Eu = 1

2(∇u+∇Tu) =

(Eiju)i,j=1,...,d its symmetric gradient. For a d2× d2 symmetric matrix A = (Aklij )

di,j,k,l=1 we denote its

bilinear form

AEu : Ev =d∑

i,j,k,l=1

AklijEijuEklv.

We assume that the entries Aklij of symmetric matrix A satisfy the following assumptions:

Aklij ∈ VMO(Ω,R) ∩ L∞(Ω,R) for i, j, k, l = 1, . . . , d and

V (r) = supx∈Ω; 0<ρ≤r, i,j,k,l=1,...,d1

|Ωρ(x)|

∫Ωρ(x)

∣∣Aklij (y)− (Akl

ij )Ωρ(x)

∣∣ dy → 0 for r → 0+; (2)

‖A‖L∞(Ω) ≤ Λ, (3)

A satisfies a suitable ellipticity condition, i.e. there exists positive constant λ so that

λ|ξ|2 ≤ A(x)ξ : ξ (4)

2

for any symmetric d × d matrix ξ and for almost all x ∈ Ω. For simplicity we assume that f ∈L2(Ω,Rd2

) is symmetric.

By a weak solution of (1) we will understand a couple (u, p) ∈ W 1,2div (Ω,Rd)× L2(Ω,R) such that∫

Ω

AEu : Ev dx−∫Ω

p div v dx =

∫Ω

f : Ev dx, ∀ v ∈ W 1,20 (Ω,Rd).

Lemma 1. (Caccioppoli inequality) (See [7] Thm. 1.1 or [10] Lemma 3.0.5)Let the coefficients A satisfy (3) and (4). Then there exist positive constants c1, c2 so that if (u, p)is a weak solution of (1) with a right hand side f ∈ L2(Ω,Rd), x ∈ Ω and 0 < ρ < R < dist(x, ∂Ω)it holds ∫

Bρ(x)

|Eu|2 dy ≤ c1(R− ρ)2

∫BR(x)

|u− (u)R|2 dy + c2

∫BR(x)

|f |2 dy. (5)

Here and in what follows we abbreviate (u)BR(x) (in case where the dependence on the center x ofBR(x) is clear) by (u)R.

Proof is a slight modification of the proof of Theorem 1.1 in [7].

Lemma 2. (Campanato inequality) (See [7], Prop. 1.9)

Let the constant coefficients A satisfy (3), (4). Then there exists a positive constant C so that forany weak solution (u, p) of (1) for f = 0 and 0 < ρ < R < dist(x; ∂Ω) it holds∫

B%(x)

|Eu|2 dy ≤ C( %R

)d∫

BR(x)

|Eu|2 dy. (6)

NotationFor a function u ∈ W 1,2(BR(x),Rd) denote by PRu(y) = PBR(x)u(y) = (u)R+SR(y−x) its orthogonalprojection on the set of rigid motions in Rd, i.e.

(SR)ij =

∫BR(x)

((u(y)− (u)R)i(y − x)j − (u(y)− (u)R)j(y − x)i) dy∫BR(x)

((y − x)2

i + (y − x)2j

)dy

.

Lemma 3. (Local Korn inequalities) (see [1])Let PR be the orthogonal projection of W 1,2(BR(x),Rd) on the set of rigid motions in Rd. Then itholds

‖PRu‖L2(BR(x),Rd) ≤ C1‖u‖L2(BR(x),Rd), (7)

‖u− PRu‖Lq(BR(x),Rd) ≤ C2R1+(d/q)−(d/p)‖Eu‖Lp(BR(x),Rd), (8)

‖∇u‖L2(BR(x),Rd) ≤C3

R‖u− (u)R‖L2(BR(x),Rd) + C4‖Eu‖L2(BR(x),Rd). (9)

3

for any 1 ≤ p < d, 1 ≤ q ≤ p∗ = pdd−p

with Cj for j = 1, . . . , 4 independent of R and x.

Definition.(Interior A-harmonic function)Let the constant coefficients matrix A satisfy (3),(4), let Ω be a Lipschitz domain. Function h ∈W 1,2

div (Ω,Rd) is called A-harmonic if∫Ω

AEh : Ev dx = 0 ∀ v ∈ W 1,20,div(Ω,R

d).

Lemma 4. (Interior A-harmonic approximation)Let 0 < λ < Λ and d ∈ N with d ≥ 2 be fixed. Then for any ε > 0 there exists a constantk(ε) = k(d, λ,Λ, ε) > 0 such that the following holds: for any bilinear form A on Rd2

satisfyingconditions (3),(4) with λ,Λ and for any u ∈ W 1,2

div (BR(x),Rd) there exists an A-harmonic functionh ∈ W 1,2

div (BR(x),Rd) such that ∫BR(x)

|Eh|2 dy ≤∫

BR(x)

|Eu|2 dy (10)

and, moreover, there exists ϕ ∈ C∞0,div

(BR(x),Rd

)such that

‖Eϕ‖L∞(BR(x),Rd) ≤1

R(11)

and ∫BR(x)

|u− h|2 dy ≤ εR2

∫BR(x)

|Eu|2 dy + k(ε)R4−d

∫BR(x)

AEu : Eϕdy

2

. (12)

Proof. First observe that it is enough to prove the lemma for x = 0 and R = 1 since the fullstatement follows by a standard homotopy argument. Denote B1(0) = B. Next, the important partof the proof is an analogue of Lemma 2.1 of [5] which in our case reads as follows:

Assertion 1Let 0 < λ < Λ and d ∈ N, with d ≥ 2, be fixed. Then for any ε > 0 there exists a constantδ(ε) = δ(d, λ,Λ, ε) ∈]0, 1] with the following property: for any bilinear form A on Rd2

satisfying theconditions (3) and (4) with λ, Λ and for any w ∈ W 1,2

div (B,Rd) such that∫B

|Ew|2 dx ≤ 1 (13)

and ∣∣∣∣∣∣∫B

AEw : Eϕdx

∣∣∣∣∣∣ ≤ δ(ε) supy∈B

|Eϕ(y)| (14)

4

for all ϕ ∈ C10,div(B,Rd), there exists an A-harmonic function g ∈ W 1,2

div (B,Rd) such that∫B

|Eg|2 dx ≤ 1 (15)

and ∫B

|w − g|2 dx ≤ ε. (16)

Let us prove now Lemma 4 from the Assertion 1.First, fix any ε > 0 take a δ = δ(ε) as in the quoted Assertion and set

w := (u− P1(u))

∫B

|Eu|2 dx

−1/2

.

Then (13) holds. (1) Assume that for w the assumption (14) holds. Then there is an A-harmonic func-tion g satisfying requirements (15),(16) and thus the function h(y) ≡ (g(y)−P1(u)(y))(

∫B|Eu|2 dz)1/2

fulfills (10). Choosing ϕ ≡ 0 we get also (11) and (12).If, vice versa, there is a nonzero ψ ∈ C1

0,div(B,RN) such that∣∣∣∣∣∣∫B

AEw : Eψ dx

∣∣∣∣∣∣ > δ supy∈B|Eψ(y)|,

we set

ϕ =ψ

supy∈B |Eψ|.

Then

1

δ

∣∣∣∣∣∣∫B

AEw : Eϕdx

∣∣∣∣∣∣ > 1, (17)

For h = P1(u) we have from Lemma 3, formula (8) and (17)

∫B

|u− P1(u)|2 dx ≤ C2

∫B

|Eu|2 dx ≤ C2

δ2

∣∣∣∣∣∣∫B

AEu : Eϕdx

∣∣∣∣∣∣2

which concludes the proof.

Proof of Assertion 1Were the conclusion false we could find ε > 0 such that for any δk = 1

kthere are Ak satisfying (3),(4),

uk ∈ W 1,2div (B) with

∫B|Euk|2 dy ≤ 1 such that∫

B

AkEuk : Eϕdy ≤ 1

ksup

B|Eϕ| (18)

1Note that the assertion is trivial for∫

B|Eu|2 dx = 0.

5

for every ϕ ∈ C10,div and at the same time for all Ak-harmonic functions hk ∈ W 1,2

div (B) such that∫B|Ehk|2 dy ≤ 1 it would hold ∫

B

|uk − hk|2 dy ≥ ε. (19)

1. Choice of convergent coefficientsAs the coefficient matrices are uniformly bounded we can choose a convergent subsequence andassume that Ak → A where A satisfies (3),(4).

2. Choice of weakly convergent subsequenceSet uk = uk − P1(uk). Then thanks to Lemma 3 formula (8)

‖uk‖W 1,2(B) ≤ c

and we can choose a subsequence which weakly converges to a function u in W 1,2div (B). Thus for (not

relabeled) subsequence

∇uk ∇u in L2(B)

uk → u in L2(B). (20)

3. u is A-harmonic

For an arbitrary ϕ ∈ W 1,20,div(B) we rewrite∫

B

AEu : Eϕdy =

∫B

(A− Ak)Eu : Eϕdy +

∫B

AkE(u− uk) : Eϕdy +

∫B

AkEuk : Eϕdy.

The first term on the right hand side tends to zero as Ak → A while the second term tends to zeroas AkEϕ → AEϕ strongly in L2(B) and E(uk − u) 0 in L2(B). As P1(uk) is antisymmetric itssymmetric gradient E(P1(uk)) = 0 and Euk = Euk. Hence the third term tends to zero becauseof (18) and u is A-harmonic. As all uk are divergence free the same is true for u. Moreover, asP1(uk) = 0 and P1 is a continuous linear operator we have also P1(u) = 0.

4. Auxiliary functions gk

Denote gk ∈ W 1,2div (B) the unique weak solution to the boundary value problem∫

B

AkEgk : Eϕdy = 0, ∀ϕ ∈ W 1,20,div(B)

gk − u ∈ W 1,20,div(B) (21)

Note that the solutions exist and they are uniformly bounded as Ak are uniformly W 1,20 -elliptic

because of (3), (4) and Korn inequality. Thus we can assume that a (non relabeled) subsequencetends weakly to a g in W 1,2

div (B) and by the same arguments as before we prove that g is a weaksolution of ∫

B

AEg : Eϕdy = 0, ∀ϕ ∈ W 1,20,div(B)

div g = 0 on B (22)

g − u ∈ W 1,20 (B).

6

As A satisfy (4), Korn inequality implies uniqueness of the solution to (22) and thus g = u.

5. Strong convergence of (gk) to uUsing ellipticity condition (4) we get∫

B

|E(u− gk)|2 dy ≤1

λ

∫B

AkE(u− gk) : E(u− gk) =

1

λ

∫B

AkEu : E(u− gk) dy −∫B

AkE(gk) : E(u− gk)

Now the first term on the right hand side tends to zero as AkEu → AEu strongly in L2 whileE(u−gk) 0 weakly in L2. The second term is equal to zero as u−gk is an admissible test functionin (21).

6. ContradictionFollowing the proof of Lemma 2.1 of [5] we set mk = max‖Egk‖L2(B), 1 and Gk = 1

mkgk. As

Egk → Eu in L2 and ‖Eu‖L2(B) ≤ lim inf ‖Euk‖L2(B) = lim inf ‖Euk‖L2(B) ≤ 1 we have that mk → 1and,finally,

Gk − gk = gk

(1

mk

− 1

)→ 0 (23)

in W 1,2(B). Thus hk = Gk + P1uk are Ak- harmonic, ‖Ehk‖L2(B) ≤ 1 and∫B

|uk − hk|2 dx =

∫B

|uk −Gk|2 dx ≤ 2(

∫B

|uk − u|2 dx+

∫B

|u− gk|2 dx+

∫B

|gk −Gk|2 dx).

All terms on the right hand side tend to zero because of (20), strong convergence of gk to u and (23)and it contradicts (19).

In the boundary case we will use the following resultLemma 5. (Equivalent norms)There exists C(d) (which does not depend on R) such that for any u ∈ W 1,2(B+

R(x),Rd) it holds

‖u‖L2(B+R(x)) ≤ C(d)

(R‖Eu‖L2(B+

R(x)) + ‖u‖L2(ΓR(x))

). (24)

If, moreover, u = 0 on ΓR(x) then there exists a constant C(d) so that

‖∇u‖L2(B+R(x)) ≤ C(d)‖Eu‖L2(B+

R(x)). (25)

Proof.As before it is enough to prove the Lemma for x = 0, R = 1, B+ = B+

1 (0), Γ = Γ1(0) and usehomotopy argument. Were the conclusion false we could find a sequence (uk) ⊂ W 1,2(B+,Rd) sothat 1 = ‖uk‖L2(B+) > k(‖Euk‖L2(B+) + ‖uk‖L2(Γ)). By Korn inequality (Lemma 3 formula (9)) (uk)is bounded in W 1,2(B+) and we can extract a (not relabeled) subsequence so that (uk) convergesweakly to u in W 1,2(B+) and, at the same time, uk → u;Euk → Eu = 0 strongly in L2(B+,Rd) and

7

uk → 0 in L2(Γ,Rd). It implies that Eddu = 12(∂ud

∂xd) = 0 on B+ and ud = 0 on Γ. Hence ud = 0 on

B+. Thus also Eid(u) = 12( ∂ui

∂xd) = 0 on B+ for any i = 1, . . . , d − 1 which, together with boundary

condition ui = 0 on Γ implies that ui = 0 on B+. Thus u is identically zero which contradicts thefact that limk→∞ ‖uk‖L2(B+ = ‖u‖L2(B+ = 1. The second assertion follows from Lemma 3 and (24).

Lemma 6. (Boundary A-harmonic approximation)Let 0 < λ < Λ and d ∈ N, with d ≥ 2, be fixed. Then for any ε > 0 there exists a constantC(ε) = C(d, λ,Λ, ε) > 0 such that the following holds: for any bilinear form A on Rd2

satisfying theconditions (3) and (4) with λ, Λ and for any u ∈ W 1,2(B+

R(x),Rd) such that u = 0 on ΓR(x) thereexists an A-harmonic h ∈ W 1,2

div (B+R(x),Rd) such that:

h = 0 on ΓR(x),∫B+

R(x)

|Eh|2 dx ≤∫

B+R(x)

|Eu|2 dx, (26)

and, moreover, there exists ϕ ∈ C10,div(B

+R(x),Rd) so that

‖Eϕ‖L∞(B+R(x)) ≤

1

R(27)

and

∫B+

R(x)

|u− h|2 dx ≤C(ε)

R4−d

∫B+

R(x)

AEuEϕdx

2

+R2

∫B+

R(x)

|div u|2 dx

+ εR2

∫B+

R(x)

|Eu|2 dx . (28)

Proof.As in the previous Lemma we will present the proof for x = 0, R = 1 with the notation B+ = B+

1 (0),Γ = Γ1(0). Lemma 6 is an easy consequence of the following assertion which was inspired by theoriginal formulation of the lemma due to J. Grotowski ([9]).

Assertion 2.Let 0 < λ < Λ and d ∈ N, with d ≥ 2, be fixed. Then for any ε > 0 there exists a

δ(ε) = δ(d, λ,Λ, ε) > 0

such that the following holds: for any bilinear form A on Rd2, satisfying the conditions (3) and (4)

8

with λ, Λ and for any u ∈ W 1,2(B+,Rd), u = 0 on Γ such that∫B+

|Eu|2 dx ≤ 1 (29)

∣∣∣∣∣∣∫

B+

AEuEϕdx

∣∣∣∣∣∣ ≤ δ(ε) supy∈B+

|Eϕ(y)| for all ϕ ∈ C10,div(B

+) (30)

‖div u‖L2(B+) ≤ δ(ε) (31)

there exists an A-harmonic h ∈ W 1,2div (B+,Rd) such that:

h = 0 on Γ,∫B+

|Eh|2 dx ≤ 1 (32)

and, moreover, ∫B+

|u− h|2 dx ≤ ε. (33)

Proof of Lemma 6 from Assertion 2First, fix an arbitrary ε and take a δ(ε) as in Assertion 2. For u from Lemma 6 set

w := u

∫B+

|Eu|2 dx

−1/2

Then w satisfies (29). If, moreover, also (30) and (31) are satisfied then there exists an A-harmonic

g ∈ W 1,2div (B+,Rd) satisfying (32),(33). Thus h = g

(∫B+ |Eu|2 dx

) 12 satisfies (26) and choosing ϕ = 0

we get also (28).If, viceversa, one of the conditions (30),(31) does not hold we proceed as in the proof of Lemma 4from Assertion 1. If (30) is not valid we repeat the arguments of the proof of Lemma 4. If (31) isnot satisfied then ∫

B+

|Eu|2 dx

1/2

≤ 1

δ(ε)‖divu‖L2(B+)

and, choosing h = ϕ = 0 we get by (24)∫B+

|u|2dx ≤ C2(d)

∫B+

|Eu|2 dx ≤ C2(d)

δ(ε)‖div u‖2

L2(B+)

and (28) is proved with C(ε) = C2(d)δ(ε)

.

Proof of Assertion 2

9

We will prove the assertion for x = 0, R = 1 and B+, Γ as in the previous proofs. Were the conclusionfalse we could find ε > 0 such that for any δk = 1

kthere are Ak satisfying (3), (4), uk ∈ W 1,2(B+,Rd)

with∫

B+ |Euk|2 dy ≤ 1, uk = 0 on Γ such that

‖div uk‖L2(B+) ≤1

k, (34)∣∣∣∣∣∣

∫B+

AkEuk : Eϕdy

∣∣∣∣∣∣ ≤ 1

ksupB+

|Eϕ| (35)

for every ϕ ∈ C10,div(B

+,Rd) and at the same time for all Ak-harmonic functions hk ∈ W 1,2div (B+,Rd)

such that∫

B+ |Ehk|2 dy ≤ 1, hk = 0 on Γ it would hold∫B+

|uk − hk|2 dy ≥ ε. (36)

The proof consists of several steps.

1.Choice of convergent coefficients.As the coefficients matrices are uniformly bounded we can choose a convergent subsequence andassume that Ak → A where A satisfies (3), (4).

2. Choice of weakly convergent subsequence of (uk)Thanks to (8), (9), Lemma 5 and zero traces of uk on Γ we have

‖uk‖W 1,2(B+) ≤ C(d)‖Euk‖L2(B+) (37)

and we can choose a subsequence which weakly converges to a function u in W 1,2(B+,Rd). Thus for(not relabeled) subsequence

∇uk ∇u in L2(B+,Rd2

),

uk → u in L2(B+,Rd).

Taking into account the compactness of trace operator, continuity of linear operator div u, and lowernorm semicontinuity we obtain

uk → u in L2(Γ,Rd),

div u = 0 on B+,∫B+

|Eu|2 dy ≤ 1.

3. The proof that u is A-harmonic copies the corresponding part of the proof of Assertion 1.

4. Auxiliary functions vk

10

Denote vk ∈ W 1,2div (B+,Rd) the unique weak solution to the boundary value problem∫

B+

AkEvk : Eϕdy = 0, ∀ϕ ∈ W 1,20,div(B

+,Rd)

vk − u ∈ W 1,20 (B+,Rd) (38)

Note that the solutions exist and they are uniformly bounded as Ak are uniformly W 1,20 -elliptic

because of (3), (4) and Korn inequality. Thus we can assume that a (non relabeled) subsequence(vk) tends to a v ∈ W 1,2(B+,Rd) weakly and by the same arguments as before we prove that v is aweak solution of ∫

B+

AEv : Eϕdy = 0, ∀ϕ ∈ W 1,20,div(B

+,Rd) (39)

div v = 0 on B+

v − u ∈ W 1,20 (B+,Rd).

As A satisfy (4), Korn inequality implies uniqueness of the solution to (39) and thus v = u.

5. Strong convergence of vk to uUsing ellipticity condition (4) we get∫

B+

|E(u− vk)|2 dy ≤1

λ

∫B+

AkE(u− vk) : E(u− vk) =

1

λ

∫B+

AkEu : E(u− vk) dy −∫

B+

AkE(vk) : E(u− vk)

Now the first term on the right hand side tends to zero as AkEu → AEu strongly in L2 whileE(u−vk) 0 weakly in L2. The second term is equal to zero as u−vk is an admissible test functionin (38). Strong convergence of vk to u follows as in (37).

6. ContradictionAs in the proof of Assertion 1 or Lemma 2.1 of [5] we set mk = max‖Evk‖L2(B+,Rd2 ), 1 and Vk =1

mkvk. Then mk → 1 and Vk − vk → 0 in W 1,2(B+,Rd) and the proof can be accomplished as the

proof of Assertion 1.

Lemma 7 (see [13], Ch. 2, Lemma 2.1.1) Let Ω be a domain in Rd, Ω0 a nonempty subdomain of Ω.Then there is a constant K such that for any F ∈ W−1,2(Ω,Rd) satisfying the condition [F, ϕ] = 0for all ϕ ∈ W 1,2

0,div(Ω,Rd) there is unique q ∈ L2(Ω,R) for which

∇ q = F,

∫Ω0

q = 0, ‖q‖L2(Ω,R) ≤ K‖F‖W−1,2(Ω,Rd).

11

3 Interior regularity.

Theorem 1. Let A satisfies condition (2), (3), (4). For any Ω′ ⊂⊂ Ω there exists positive constantC(Ω′) such that for u, p which solves (1) with a right hand side f ∈ L2,µ(Ω,Rd), (0 < µ < d) thecouple (Eu, p) belong to L2,µ(Ω′,Rd)× L2,µ(Ω′,R) and the inequality

‖Eu‖L2,µ(Ω′,Rd) + ‖p‖L2,µ(Ω′,R) ≤ C(Ω′)‖f‖L2,µ(Ω,Rd)

holds.

Proof.Without loss of generality assume that x = 0 ∈ Ω′ and 0 < % < R, 2R < dist(Ω′; ∂Ω) and write B%

and P%u instead of B%(x) and PB%(x)u. As the matrix S% is antisymmetric, E(P%u) = 0, div(P%u) = 0and for any % the couple (u − P%u, p) solves (1). Moreover, from Lemma 1 applied to u − P%u wehave ∫

B%/2

|Eu|2 dx ≤ c1%2

∫B%

|u− P%u|2 dx+ c2

∫B%

|f |2 dx. (40)

We set A = (A)R. By means of Lemma 4 there exists an A-harmonic h ∈ W 1,2div (BR(x),Rd) such that

(10), (11) and (12) hold. Now from (40) we get

∫B%/2

|Eu|2 dx ≤ 2c1%2

∫B%

|h− P%h|2 dx+

∫B%

|u− P%u− (h− P%h)|2 dx

+ c2

∫B%

|f |2 dx. (41)

The function h− P%h is A-harmonic and from Lemma 3 and Lemma 2 it follows

I1 ≡∫B%

|h− P%h|2 dx ≤ d1%2

∫B%

|Eh|2 dx ≤ d2%2( %R

)d∫

BR

|Eh|2 dx, (42)

where d2 = d1C. Hence from (10) we have

I1 ≤ d2%2( %R

)d∫

BR

|Eu|2 dx. (43)

Further by means of (7) and (12)

I2 ≡∫B%

|u− P%u− (h− P%h)|2 dx ≤ d3

∫B%

|u− h|2 dx

≤ d3

εR2

∫BR

|Eu|2 dx+ k(ε)R4−d

∫BR

AEu : Eϕdx

2 . (44)

12

for a suitable function ϕ from (11). The function u solves Stokes system (1) and we have∫BR

AEu : Eϕdx =

∫BR

[A− A

]Eu : Eϕdx+

∫BR

fEϕdx, (45)

i.e. ∣∣∣∣∣∣∫

BR

AEu : Eϕdx

∣∣∣∣∣∣2

≤ 2

∫BR

[A− A

]Eu : Eϕdx

2

+

∫BR

fEϕdx

2 (46)

hence by means of (11) and Holder inequality we have∣∣∣∣∣∣∫

BR

AEu : Eϕdx

∣∣∣∣∣∣2

≤ 2

R2

∫BR

∣∣A− A∣∣2 dx ∫

BR

|Eu|2 dx+Rd

∫BR

|f |2 dx

(47)

and as A, A satisfy (2), (3) also ∫BR

∣∣A− A∣∣2 dx ≤ 2RdΛV (R)

Using preceding inequality in (44) we get

I2 ≤ d3R2

(ε+ 4k(ε)ΛV (R))

∫BR

|Eu|2 dx+ 2k(ε)

∫BR

|f |2 dx

. (48)

Inserting (43) and (48) into (41) we achieve the following estimate∫B%/2

|Eu|2 dx ≤

[2c1d2

( %R

)d

+ 2c1d3

(R

%

)2

(ε+ 4Λk(ε)V (R))

] ∫BR

|Eu|2 dx

+

(c2 + 4c1d3k(ε)

(R

%

)2)‖f‖2

L2(BR,Rd). (49)

The above inequality can be rewritten as

φ(%

2

)≤

[A( %R

)d

+ εo

(R

%

)2]φ(R) + B

(R

%

)Rµ,

where

φ(R) =

∫BR

|Eu|2 dx (50)

A = 2c1d2, εo = 2c1d3 (ε+ 4Λk(ε)V (R)) , B(R

%

)=

(c2 + 4c1d3k(ε)

(R

%

)2)‖f‖2

L2,µ(Ω,Rd). (51)

13

For any τ ∈ (0, 1/2), 0 < R < 1/2 dist(x, ∂Ω), % = 2τR and α ∈ (µ, d) we have

φ(τR) ≤ τα[A2dτ d−α + εoτ−2−α]φ(R) + B(1/τ)Rµ.

Take τ so small that A2dτ d−α ≤ 1/2. For such τ we can choose subsequently ε < τ2+α

8c1d3and Ro so

small that V (Ro) <τ2+α

16c1d3Λk(ε). Hence also εoτ

−2−α < 1/2. Without loss of generality we can assume

that Ro < 1/2 dist(x, ∂Ω). Thus for 0 < R < Ro we get

φ(τR) ≤ ταφ(R) + B(τ)Rµ

and the proof continues as the proof of Lemma 0.6 of [7] to the final estimate

φ(%) ≤(%

Ro

)µ

[φ(Ro) + B(τ)Rµo ] .

Then ∫B%(x)

|Eu(y)|2 dy ≤ C(Ro)%µ||f ||L2,µ(BRo (x),Rd)

To estimate the pressure term we use Lemma 7 with Ω0 = Ω = B%(x) and the functional F definedby

[F, ϕ] =

∫B%(x)

A(x)Eu : Eϕ−∫

B%(x)

fEϕ.

We have [F, ϕ] = 0 for all ϕ ∈ W 1,20,div(B%(x),Rd), F ∈ W−1,2(B%(x),Rd) and, moreover,

|[F, ϕ]| ≤ (Λ‖Eu‖L2(B%(x),Rd) + ‖f‖L2(B%(x),Rd)).‖Eϕ‖L2(B%(x),Rd),

‖F‖W−1,2 = supϕ∈W 1,2

0 (B%(x),Rd),ϕ 6=0

|[F, ϕ]|‖ϕ‖W 1,2

0 (B%(x),Rd)

≤ c(‖Eu‖L2(B%(x),Rd) + ‖f‖L2(B%(x),Rd)

).

According to Lemma 7 there is a unique q = p− (p)% so that ∇ q = F ,∫

B%(x)

q = 0 and

∫B%(x)

|p− (p)ρ|2 ≤ K2‖F‖2W−1,2 ≤ c

∫B%(x)

|Eu|2 +

∫B%(x)

|f |2

≤ c%µ.

As µ < d the equivalence of Morrey and Campanato spaces concludes the proof of the inequality∫BR(0)

(|Eu(y)|2 + |p(y)|2

)dy ≤ C(Ω′)Rµ‖f‖L2,µ(Ω,Rd). (52)

Let C1 be a positive constant such that

‖u‖L2,µ(Ω′) ≤ C1 supx∈Ω,R<min1/2dist(Ω′,∂Ω),Ro

R−µ

∫BR(x)

|u|2dx.

Thus, using (52), we have the assertion with C = C1C(Ω′) where C(Ω′) is the constant of (52)

14

4 Boundary regularity.

In this Section we deal with system (1)

−divAEu+∇ p = −divf

div u = 0

on a domain Ω ∩ BR(x) where x ∈ ∂Ω. Fix a positive sufficiently small δ ∈ (0, 1/2).

Since Ω is of the class C1 and bounded (see e.g. [11] pag. 305) for arbitrarily small R > 0 there existsan open finite covering BR(xj)j=1,...,ν of ∂Ω by open balls BR(xj) and for each j a suitable systemof coordinates y1, . . . , yd composed of translations and rotations with the origin Oj at xj. Withoutloss of generality we can suppose that the new system of coordinates is such that the hyperplanetangent to ∂Ω at xj has equation yd = 0. Recall that if we accompany the translation and rotation ofindependent variables by the corresponding transform of dependent variables both the divergence oftransformed functions and the form and the ellipticity constant of the system (1) remain unchanged.For these reasons we keep the notation of u, p and of the coefficients. For the sake of simplicity wewill drop the index j relative to the coordinate system.Moreover,in this system there exists a C1–function ζ defined on a domain G ⊂ Rd−1 containing theorigin O′ of Rd−1 such that

(a) the set ∂Ω ∩ BR(x) can be represented by an equation of the type:

yd = ζ(y1, . . . , yd−1); y′ = [y1, ..., yd−1] ∈ G,

(b) each x ∈ Ω ∩ BR(x) satisfies

yd > ζ(y1, . . . , yd−1); y′ = [y1, ..., yd−1] ∈ G,

(c)ζ(O′) = ∇ζ(O′) = 0 and sup

G|∇ζ| < δ < 1/2.

For such domains the portion of boundary within the ball BR(O) can be straightened by means ofthe smooth transformation

ψi(y) = yi, i = 1, 2, . . . , d− 1

ψd(y) = yd − ζ(y1, . . . , yd−1). (53)

It turns out that z = ψ(y) = (ψ1(y), . . . , ψd(y)) is a C1–diffeomorphism with Jacobian J = 1 verifyingthe following properties (see e.g. [11] p. 305):

(i) 12|y| ≤ |ψ(y)| ≤ 3

2|y|, ∀y ∈ BR(O) ∩ Ω,

(ii) z ∈ Rd : zd = 0, |z′| < 12R ⊂ ψ (BR(O) ∩ ∂Ω) ⊂ z ∈ Rd : zd = 0, |z′| < 3

2R,

(iii) B+R2

(0) ⊂ ψ (BR(O) ∩ Ω) ⊂ B+32R(0), B 2

3R(O) ∩ Ω ⊂ ψ−1

(B+

R(0))⊂ B2R(O) ∩ Ω.

15

If z = ψ(y) ∈ B+R(0) we set

u(z) = u(ψ−1(z)), p(z) = p(ψ−1(z)),

B(z) = A(ψ−1(z)), gi(z) = fi(ψ−1(z)).

(54)

In what follows partial derivatives and differential operators when applied to functions u, p, ϕ concerny-variables and when applied to functions u, p, ϕ concern z-variables.

Eiju = 1/2(∂ui

∂yj

+∂uj

∂yi

) = 1/2(∂ui

∂zj

+∂uj

∂zi

)− 1/2(∂ui

∂zd

∂ζ

∂zj

+∂uj

∂zd

∂ζ

∂zi

) ≡ Eiju+ εij(u)

divu = divu+∂ui

∂zd

∂ζ

∂zi

≡ divu+ η(u).

Thus for suitably small R the change of variables in the system (1) on B2R(O) ∩ Ω yields

u ∈ W 1,2(B+R(0),Rd),

u = 0 on ΓR(0),

divu+ η(u) = 0,∫B+

R(0)

BEu : Eϕdz +

∫B+

R(0)

(p− π) divϕdz =

∫B+

R(0)

f : (Eϕ+ ε(ϕ)) dz +

∫B+

R(0)

H(u, p− π, ϕ) dz,

∀ϕ ∈ W 1,20 (B+

R(0),Rd). (55)

HereH(u, p− π, ϕ) = −B(Eu : ε(ϕ) + ε(u) : Eϕ+ ε(u) : ε(ϕ))− (p− π) η(ϕ).

(It is obvious that if a couple (u, p) solves (1) then for any constant π also (u, p− π) is a solution.)Further as A satisfies (2),(3),(4) on Ω the symmetric matrix B satisfies the same conditions on B+

R(0)and its elements belong to VMO with uniformly bounded VMO-modulus V (r).

First we formulate the corresponding Caccioppoli inequality and outline the proof.

Lemma 8. (Caccioppoli inequality) Assume (3),(4). There exist positive constants c1, c2 sothat if u, p is a weak solution of (55) with a right hand side f ∈ L2(B+

R(0),Rd), zo ∈ ΓR(0) and0 < ρ < r < R− |zo| it holds∫

B+ρ (zo)

(|Eu|2 + |p− (p)ρ|2

)dz ≤ c1

(r − ρ)2

∫B+

r (zo)

|u|2 dz + c2

∫B+

r (zo)

|f |2 dz. (56)

Sketch of the proofPart 1. Without loss of generality assume that λ ≤ 1. Choose in (55) test function ϕ = uτ 2, where τis a smooth function supported in Br(zo), τ ≡ 1 on Bρ(zo) and |∇τ | ≤ C

r−ρon Br(zo). Using (3),(4)

16

we get

λ

∫B+

r (zo)

|Eu|2τ 2 ≤∫

B+r (zo)

|B||Eu||u||τ ||∇τ |+∫

B+r (zo)

|p− π|(|η(u)|τ 2 + 2|u||τ ||∇τ |)

+

∫B+

r (zo)

|f ||Eϕ+ ε(ϕ)|+∫

B+r (zo)

H(u, p− π, ϕ)

= I1 + I2 + I3 + I4. (57)

By standard use of Holder and Young inequalities we obtain that for any positive σ there is a positiveC(σ) such that

I1 ≤ σ

∫B+

r (zo)

|Eu|2τ 2 +C(σ)

(r − ρ)2

∫B+

r (zo)

|u|2,

I2 ≤ σ

∫B+

r (zo)

|p− π|2τ 2 +C(σ)

(r − ρ)2

∫B+

r (zo)

|u|2 + δ2C(σ)

∫B+

r (zo)

|∇u|2τ 2,

I3 ≤ σ

∫B+

r (zo)

|Eu|2τ 2 + δ2σ

∫B+

r (zo)

|∇u|2τ 2 +C(σ)

(r − ρ)2

∫B+

r (zo)

|u|2 + C(σ)

∫B+

r (zo)

|f |2,

I4 ≤ 2σ

∫B+

r (zo)

|Eu|2τ 2 + 2σ

∫B+

r (zo)

|p− π|2τ 2 + δ2C(σ)

∫B+

r (zo)

|∇u|2τ 2 +C(σ)

(r − ρ)2

∫B+

r (zo)

|u|2).

Hence for σ < λ8

we use the properties of τ and Lemma 5 to get

1

2λ

∫B+

ρ (zo)

|Eu|2 ≤ 3σ

∫B+

r (zo)

|p−π|2 + δ2C(d)C(σ)

∫B+

r (zo)

|Eu|2 +C(σ)

(r − ρ)2

∫B+

r (zo)

|u|2 +C(σ)

∫B+

r (zo)

|f |2.

(58)

Part 2. Choose now the test function ϕ as a solution of the problem

divϕ = p− (p)r on B+r (zo)

ϕ = 0 on ∂B+r (zo).

Observe that as∫

B+r (zo)

(p−(p)r)dx = 0 there exists a solution ϕ and a positive constant C independent

of r so that

||∇ϕ||2 ≤ C||p− (p)r||2. (59)

Choosing π = (p)r in (55) it holds∫B+

r (zo)

(p− (p)r)divϕ =

∫B+

r (zo)

|p− (p)r|2

17

and

|∫

B+r (zo)

BEuEϕ| ≤ C(σ)

∫B+

r (zo)

|Eu|2 + σ

∫B+

r (zo)

|p− (p)r|2

|∫

B+r (zo)

f : (Eϕ+ ε(ϕ))| ≤ C(σ)

∫B+

r (zo)

|f |2 + σ

∫B+

r (zo)

|p− (p)r|2

|∫

B+r (zo)

H(u, p, ϕ)| ≤ δC(σ)

∫B+

r (zo)

|∇u|2 + δ(σ + C)

∫B+

r (zo)

|p− (p)r|2.

Thus Lemma 5 implies∫B+

r (zo)

|p− (p)r|2 ≤ (2σ + δ(C + σ))

∫B+

r (zo)

|p− (p)r|2

+ C(σ)(1 + δC(d))

∫B+

r (zo)

|Eu|2 + C(σ)

∫B+

r (zo)

|f |2. (60)

Choosing σ = 14, δ < 1

4(C+1)we get

∫B+

r (zo)

|p− (p)r|2 ≤ C1

∫B+

r (zo)

|Eu|2 + C2

∫B+

r (zo)

|f |2, (61)

where C1, C2 depend on now fixed σ and C(σ). Substituting (61) into (58) yields

1

2λ

∫B+

ρ (zo)

|Eu|2 ≤ (3σC1 + δ2C(d)C(σ))

∫B+

r (zo)

|Eu|2 +C(σ)

(r − ρ)2

∫B+

r (zo)

|u|2 + (C(σ) + C2)

∫B+

r (zo)

|f |2.

(62)

The choice of σ and δ so small that 23σC1+δ2C(d)C(σ)

λ≤ ε0 where ε0 is a constant from Lemma 0.5 of

[7] makes all the assumptions of this lemma satisfied and from the assertion of this lemma we finallyobtain also the assertion of Lemma 8.

In what follows we use the analogy of Theorem 1 for system (55). As the proof are very similar wepresent only the boundary version here.

Proposition Let B satisfies condition (2), (3) and (4). Then there exist positive constants C andR0 such that for any u, p which solves (55) with right hand side f ∈ L2,µ(B+

R(0),Rd), (0 < µ < d),

it holds ∫B+

R(zo)

(|Eu(z)|2 + |p(z)− (p)R|2

)dz ≤ CRµ‖f‖2

L2,µ(B+

R(0),Rd)

(63)

for any zo ∈ ΓR/2(0) and 0 < R < minR0, R/4.

18

Proof.First, fix zo ∈ ΓR/2(0), R ∈ (0, R/4) and 0 < % < R. From Lemma 8 we have∫

B+%/2

(zo)

|Eu|2 dz ≤ c1%2

∫B+

% (zo)

|u|2 dz + c2

∫B+

% (zo)

|f |2 dz. (64)

Denote B the integral mean value of the coefficients B over B+R(zo). By means of Lemma 6 there

exists h ∈ W 1,2div (B+

R(zo),Rd) such that (26), (27) and (28) hold. Now from (64) we get

∫B+

%/2(zo)

|Eu|2 dz ≤ 2c1%2

∫B+

% (zo)

|h|2 dz +

∫B+

% (zo)

∣∣u− h∣∣2 dz

+ c2

∫B+

% (zo)

|f |2 dz. (65)

The function h is B-harmonic and from Poincare inequality, Lemma 5, boundary analogy of Cam-panato Lemma 2 and (26) it follows

I1 ≡∫

B+% (zo)

∣∣h∣∣2 dz ≤ d1%2

∫B+

% (zo)

|Eh|2 dz ≤ d2%2( %R

)d∫

B+R(zo)

|Eh|2 dz

≤ d2%2( %R

)d∫

B+R(zo)

|Eu|2 dz. (66)

Further by means of (28)

I2 ≡∫

B+% (zo)

∣∣u− h∣∣2 dz ≤ εR2

∫B+

R(zo)

|Eu|2 dz+

k(ε)

R4−d

∫B+

R(zo)

BEu : Eϕdz

2

+R2

∫B+

R(zo)

|η(u)|2 dz

(67)

for a suitable function ϕ from (27). The function u solves Stokes system (55) and we have∫B+

R(zo)

BEu : Eϕdz =

∫B+

R(zo)

[B −B

]Eu : Eϕdz +

∫B+

R(zo)

(f(Eϕ+ ε(ϕ)) +H(u, p− (p)R, ϕ)) dz

i.e. ∣∣∣∣∣∣∣∫

B+R(zo)

BEu : Eϕdz

∣∣∣∣∣∣∣2

≤ 2

∫B+

R(zo)

[B −B

]Eu : Eϕdz

2

+

2

∫B+

R(zo)

(f(Eϕ+ ε(ϕ)) +H(u, p− (p)R, ϕ)) dz

2

.

19

Hence by means of (27), Holder inequality, (25) and (61) we have∣∣∣∣∣∣∣∫

B+R(zo)

BEu : Eϕdz

∣∣∣∣∣∣∣2

≤ C

R2

∫B+

R(zo)

∣∣B −B∣∣2 dz ∫

B+R(y)

|Eu|2 dz +Rd

∫B+

R(zo)

(|f |2 + δ2|Eu|2) dz

(68)

and as B, B satisfy (2), (3) also ∫B+

R(zo)

∣∣B −B∣∣2 dz ≤ 2RdΛV (R)

Using preceding inequality in (68) we get

I2 ≤ R2

(ε+ 4k(ε)ΛV (R) + Cδ2) ∫B+

R(zo)

|Eu|2 dz +(2k(ε) + Cδ2

) ∫B+

R(zo)

|f |2 dz

. (69)

Inserting (66) and (69) into (65) we achieve the following estimate∫B+

%/2(zo)

|Eu|2 dz ≤

[2c1d2

( %R

)d

+ 2c1

(R

%

)2 (ε+ 4Λk(ε)V (R) + Cδ2

)] ∫B+

R(zo)

|Eu|2 dz

+

(c2 + 2c1

(R

%

)2)(

2k(ε) + Cδ2)‖f‖2

L2(B+R(zo),Rd)

. (70)

The above inequality can be rewritten as

φ(%

2

)≤

[A( %R

)d

+ εo

(R

%

)2]φ(R) + B

(R

%

)Rµ (71)

where

φ(R) =∫

B+R(zo)

|Eu|2 dz

A = 2c1d2, εo = 2c1(ε+ 4Λk(ε)V (R) + Cδ2

),

B(R

%

)=

(c2 + 2c1

(R

%

)2)(

2k(ε) + Cδ2)‖f‖2

L2(B+R(zo),Rd)

and the proof can be finished as the proof of Theorem 1.

Theorem 2.Let A satisfies condition (2), (3), (4). Then there exists a positive constant C such that for any u,p which solves (1) on Ω with a right hand side f ∈ L2,µ(Ω,Rd), (0 < µ < d), it holds

‖Eu‖L2,µ(Ω) + ‖p‖L2,µ(Ω) ≤ C(R0)‖f‖L2,µ(Ω).

20

The proof follows in by the standard partition of unity technique from Theorem 1 and Proposition.

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21