This section is devoted to the proofs related to the Bingham problem.
4.1 Preliminaries
We require a local variation of Lemma 2.2. Recall that Cϕ (resp.,Cϕ,n andCn) denote generic constants whose value exclusively depends on ϕ(resp., on ϕ, nand on n).
Lemma 4.1 (local estimate) Given a bounded open set Ω ⊂ Rn, u ∈ H1(Ω)n, ϕ ∈ Cc∞(Ω) withϕ≥0, and a positiveε≤2−4kϕ∇ϕk−2L∞(Ω)n, there exists a unique solution uε∈H1(Ω)n to
Z
Ω
uε·w+ε Z
Ω
ϕ4∇uε:∇w= Z
Ω
u·w for allw∈H1(Ω)n. (4.1) Moreover, ϕ2∇2uε∈L2(Ω)n×n×n, ϕ4∇3uε∈L2(Ω)n×n×n×n, and
kuεkL2(Ω)n ≤ kukL2(Ω)n, (4.2) k∇uεkL2(Ω)n×n ≤ 1
1−8εkϕ∇ϕk2L∞(Ω)n
k∇ukL2(Ω)n×n, (4.3) k∇uεk2L2(Ω)n×n+ ε
2kϕ2∆uεk2L2(Ω)n ≤ Z
Ω
∇uε:∇u+ 25εkϕ∇ϕk2L∞(Ω)nk∇uk2L2(Ω)n×n, (4.4) kϕ2∇2uεkL2(Ω)n×n×n ≤ kϕ2∆uεkL2(Ω)n+Cϕ,nkukH1(Ω)n, (4.5) kDuεkL1(Ω)n×n ≤ kDukL1(Ω)n×n+Cϕ,nε kDukL1(Ω)n×n+kϕ2∇uεkL2(Ω)n×n
. (4.6)
Proof. Since the uniqueness is obvious, the proof focuses on the existence of a solution to (4.1).
Given anyγ >0, let uεγ ∈H1(Ω)n solve Z
Ω
uεγ·w+ε Z
Ω
(ϕ4+γ)∇uεγ :∇w= Z
Ω
u·w for all w∈H1(Ω)n. (4.7) The test function w=uεγ shows
kuεγkL2(Ω)n≤ kukL2(Ω)n. (4.8) Since
uεγ−εdiv((ϕ4+γ)∇uεγ) =u in Ω (4.9) holds in the sense of distributions, we have div((ϕ4 +γ)∇uεγ) = (uεγ −u)/ε ∈ L2(Ω)n, so that (ϕ4 +γ)∇uεγ ∈ H(div,Ω)n. The comparison of (4.7) with (4.9) proves ∇uεγν = 0 in H−1/2(∂Ω)n. Since
div((ϕ4+γ)∇uεγ) = (ϕ4+γ)∆uεγ +∇uεγ∇(ϕ4) (4.10) (the Laplacian acts componentwise), we have (ϕ4+γ)∆uεγ ∈L2(Ω)n; whence ∆uεγ ∈L2(Ω)n. This and the above homogeneous Neumann boundary conditions prove
Z
Ω
∇uεγ :∇v=− Z
Ω
v·∆uεγ for allv ∈H1(Ω)n. (4.11)
The test function v=uεγ −u in (4.11) and (4.9)-(4.10) verify Z
Ω
∇uεγ : (∇uεγ− ∇u) =−ε Z
Ω
∆uεγ·
(ϕ4+γ)∆uεγ+∇uεγ∇(ϕ4)
. (4.12) The crucial point is the observation that
|∇(ϕ4)|= 4ϕ3|∇ϕ| ≤4kϕ∇ϕkL∞(Ω)nϕ2. (4.13) This, a Cauchy–Schwarz inequality, and a weighted Young inequality result in
R
Ω∆uεγ ·(∇uεγ∇(ϕ4))
≤4kϕ∇ϕkL∞(Ω)nkϕ2∆uεγkL2(Ω)nk∇uεγkL2(Ω)n×n
≤8kϕ∇ϕk2L∞(Ω)nk∇uεγk2L2(Ω)n×n+12kϕ2∆uεγk2L2(Ω)n. Its direct consequence
1 2
Z
Ω
(ϕ4+γ)|∆uεγ|2−8kϕ∇ϕk2L∞(Ω)nk∇uεγk2L2(Ω)n×n ≤ Z
Ω
∆uεγ·
(ϕ4+γ)∆uεγ+∇uεγ∇(ϕ4) is substituted in (4.12) for the proof of
1−8εkϕ∇ϕk2L∞(Ω)n
k∇uεγk2L2(Ω)n×n+ ε 2
Z
Ω
(ϕ4+γ)|∆uεγ|2 ≤ Z
Ω
∇uεγ :∇u. (4.14) Recall that ε≤2−4kϕ∇ϕk−2L∞(Ω)n and deduce
k∇uεγkL2(Ω)n×n≤ 1
1−8εkϕ∇ϕk2L∞(Ω)n
k∇ukL2(Ω)n×n ≤2k∇ukL2(Ω)n×n. (4.15) The combination of (4.14)-(4.15) reveals that
k∇uεγk2L2(Ω)n×n+ε 2
Z
Ω
(ϕ4+γ)|∆uεγ|2≤ Z
Ω
∇uεγ :∇u+25εkϕ∇ϕk2L∞(Ω)nk∇uk2L2(Ω)n×n. (4.16) The bounds (4.15)-(4.16) onuεγ imply thatuεγ is bounded inH1(Ω)nandϕ2∆uεγ is bounded in L2(Ω)nasγ →0. Hence, there exists a sequence γj →0 such that uεγj * uε weakly inH1(Ω)n andϕ2∆uεγj * ϕ2∆uεweakly inL2(Ω)n. This and the sequential weak lower semi-continuity of norms show that the weak limituε∈H1(Ω)nsolves (4.1) and, owing to (4.8) and (4.15)-(4.16), satisfies the estimates (4.2)-(4.4).
The further investigation of the unique solution uε∈H1(Ω)n to (4.1) rewrites it as
uε−εdiv(ϕ4∇uε) =u in Ω (4.17)
in the sense of distributions. Since vε =ϕ2uε∈H1(Ω)n has compact support, its extension by zero vε∈H1(Rn)n satisfies
∇vε=ϕ2∇uε+uε⊗ ∇(ϕ2)∈L2(Ω)n×n (4.18) and, in the sense of distributions inRn for any i, j= 1, . . . , n,
∂ij2vε=ϕ2∂ij2uε+∂iuε∂j(ϕ2) +∂i(ϕ2)∂juε+uε∂ij2(ϕ2). (4.19)
Recall that ϕ2∆uε∈L2(Ω)n to deduce from (4.19) that ∆vε ∈L2(Ω)n satisfies
k∆vεkL2(Ω)n ≤ kϕ2∆uεkL2(Ω)n+Cϕ,nkuεkH1(Ω)n. (4.20) Since vε has compact support in Ω, vε ∈H1(Rn)n satisfies ∆vε ∈L2(Rn)n. Invoking a Fourier transformation or the elliptic regularity of the Poisson problem on a large ball that includes Ω [15, Equ. (3,1,2,2)], we deduce that vε ∈H2(Rn)n satisfies
k∇2vεkL2(Ω)n×n×n ≤ k∆vεkL2(Ω)n. (4.21) Recall (4.2)-(4.3) withε≤2−4kϕ∇ϕk−2L∞(Ω)n to inferkuεkH1(Ω)n ≤2kukH1(Ω)n. This and (4.19) show thatϕ2∇2uε∈L2(Ω)n×n×n satisfies
kϕ2∇2uεkL2(Ω)n×n×n ≤Cϕ,nkukH1(Ω)n+k∇2vεkL2(Ω)n×n×n ≤Cϕ,nkukH1(Ω)n+kϕ2∆uεkL2(Ω)n
using (4.20)-(4.21) in the last step. This is (4.5).
Since div(ϕ4∇uε) = ϕ4∆uε+∇uε∇(ϕ4), (4.17) shows ∇(ϕ4∆uε) ∈ L2(Rn)n×n with compact support in Ω. This and the above arguments for the proof of (4.20)-(4.21) eventually lead to ϕ4∇3uε∈L2(Ω)n×n×n×n; further details on this are omitted here.
It remains to prove (4.6). Define jδ as in (2.4) but let the argumentξ be a (symmetric) matrix rather than a vector. Recall the above properties ofjδ hold in the example at hand as well. For instance, the convexity of jδ implies that
jδ(Duε)−jδ(Du)≤j0δ(Duε) : (Duε−Du). (4.22) Moreover,jδ∈C2 is convex (its Hessian is positive semi-definite) andjδas well as its derivative jδ0 are globally Lipschitz continuous with|jδ0(F)| ≤1 and|F| |jδ00(F)| ≤1 for anyF ∈Rn×n. We remark that for a symmetric matrix S and an arbitrary matrixA, one hasS:A=S :At=S : (A+At)/2. Consequently,
jδ(Duε)−jδ(Du)≤jδ0(Duε) :∇(uε−u). (4.23) Recall (4.17) to compute, for anyi, j = 1, . . . , n, the derivatives in the sense of distributions as follows:
ε−1[∇(uε−u)]ij = [∇(div(ϕ4∇uε))]ij =
n
X
k=1
∂jk2 (ϕ4∂kuiε)
=
n
X
k=1
∂k ϕ4∂jk2 uiε+∂j(ϕ4)∂kuiε
=
n
X
k=1
∂k ϕ4∂jk2 uiε+ 2∂j(ϕ4)(Duε)ki−∂j(ϕ4)∂iukε
=
n
X
k=1
∂k
ϕ4∂jk2 uiε+ 2∂j(ϕ4)(Duε)ki−δik∂j(ϕ4) tr(Duε)
−∂j(ϕ4) ∂iukε−δiktr(Duε) with the Kronecker deltaδik. Since Pn
k=1∂k(∂iukε−δiktr(Duε)) = 0, this is recast as
∇(uε−u) =ε
n
X
k=1
∂k
ϕ4∂k∇uε+ 2(Duε)k⊗ ∇(ϕ4)−tr(Duε)ek⊗ ∇(ϕ4)
−ε (∇uε)t−tr(∇uε) Id
∇2(ϕ4).
(4.24)
The integral over Ω of the combination of (4.23)-(4.24) reads
An integration by parts in the first term on the right-hand side of (4.25) shows that
n term on the right-hand side of (4.25) shows the identity below, which is followed by a Cauchy–
Schwarz inequality with respect to the symmetric and positive semi-definite Hessian ofjδ. This proves a.e. in Ω, the last term is controlled by C
ϕ,n2
The combination of (4.25)-(4.28) involves the cancellation of the upper bound (4.26) with the first term on the right-hand side of (4.27). This proves
Z and a Cauchy–Schwarz inequality over the support ofϕ concludes the proof of (4.6).
4.2 Proof of Theorem 1.6
The test functionsv= 0 and v= 2u in (1.31) provide uε to the solutionu∈H to (1.31). Recall thatuεsatisfies (4.17) andϕhas compact support, so thatuεis equal tououtside the support ofϕ; whenceuε∈H (also in case of Dirichlet boundary conditions). Thus, if u∈H01(Ω)n, we have also uε ∈H01(Ω)n. The test functionv =uε∈H in (1.31) plus elementary algebra proves
αkuε−uk2L2(Ω)n+µ|uε−u|2H1(Ω)n+ from below by (4.6), and (4.17) rewrites the integralR
Ωf·(u−uε) =−εR
Ωf·div(ϕ4∇uε). These arguments and a division byε >0 lead in (4.32) to the estimate
µ Recall the convention on the generic constants Cnϕ and estimates like (4.13) or the estimate k∇uεkL2(Ω)n×n ≤2k∇ukL2(Ω)n×n from (4.3) (for smallε) to rewrite this as
µ Cnϕ
kϕ2∆uεk2L2(Ω)n≤ kDukL1(Ω)n×n+k∇ukL2(Ω)n×n+µk∇uk2L2(Ω)n×n
+kfkL2(Ω)nkϕ2∆uεkL2(Ω)n+kfkL2(Ω)nk∇ukL2(Ω)n×n. (4.34)
Except the term kfkL2(Ω)nkϕ2∆uεkL2(Ω)n, that shall be absorbed, the a priori bounds (4.30)-(4.31) control the remaining terms on the right-hand side of (4.34). This leads to
µ Cnϕ
kϕ2∆uεk2L2(Ω)n ≤(1 α + 1
µ)kfk2L2(Ω)n+ 1
√αµkfkL2(Ω)n. (4.35) The combination with (4.5) bounds kϕ2∇2uεkL2(Ω)n×n×n independently ofε >0. Moreover, the first two terms in (4.32) have been neglected before, but lead to the control of
kϕ2∇2uεkL2(Ω)n×n×n+kuεkH1(Ω)n+α
εku−uεk2L2(Ω)n+µ
ε|u−uε|2H1(Ω)n
by a global upper bound independently ofε >0 asε→0. Hence, there exists a sequenceε→0 (not labelled explicitly) such that uε,∇uε, ϕ2∇2uε are bounded and convergence component-wise weakly in L2(Ω) to the weak limit u,∇u, ϕ2∇2u. In particular, ϕ2∇2u ∈ L2(Ω)n×n×n.
This and (4.35) lead to the assertion (1.32).
4.3 Proof of Theorem 1.7
The test functionsv= 0 and v= 2u in (1.37) provide α
Z
Ω
|u|2+ψ(u) = Z
Ω
f ·u. (4.36)
This leads to the explicit bounds on u inL2(Ω)n, namely kukL2(Ω)n ≤ kfkL2(Ω)n
α and ψ(u)≤
kfk2L2(Ω)n
α . (4.37)
Given any µ >0, let uµ ∈H solve (1.31) with eitherH =H01(Ω)n (Dirichlet) or H =H1(Ω)n (Neumann). Theorem 1.6 proves that uµ ∈ H∩Hloc2 (Ω)n satisfies (1.32) (with uµ replacing u therein) for any ϕ∈Cc∞(Ω) with ϕ≥0. Apply Lemma 4.1 as in the proof of Theorem 1.6 and define uε as an approximation to uµ (replace u in the lemma by uµ) and recall the estimate (4.33) (with uµ replacingu). Since f ∈L2(Ω)n∩Hloc1 (Ω)n, the first term in the right-hand side of (4.33) reads
− Z
Ω
f·div(ϕ4∇uε) = Z
Ω
ϕ4∇f :∇uε ≤ α
2kϕ2∇uεk2L2(Ω)n×n+ 1
2αkϕ2∇fk2L2(Ω)n×n. (4.38) Rewrite the a priori bounds (4.30)-(4.31) as
kDuµkL1(Ω)n×n+µk∇uµk2L2(Ω)n×n ≤ kfk2L2(Ω)n
α . (4.39)
Use (4.38)-(4.39) in (4.33) and follow the arguments in the proof of Theorem 1.6 to deduce α
2kϕ2∇uεk2L2(Ω)n×n ≤ 1
2αkϕ2∇fk2L2(Ω)n×n+Cnϕ kϕ2∇uεkL2(Ω)n +
kfk2L2(Ω)n α
!
. (4.40)
This andCnϕkϕ2∇uεkL2(Ω)n×n ≤Cnϕ2 /α+α4kϕ2∇uεk2L2(Ω)n×n provide theε-independent bound ε−1/2kuµ−uεkL2(Ω)+kϕ2∇uεkL2(Ω)n×n ≤ Cnϕ
α 1 +kfkL2(Ω)n+kϕ2∇fkL2(Ω)n×n
. (4.41) The first term in (4.41) stems from the first term in (4.32) and was simply suppressed in the above formulas. Given any fixed µ > 0, there exists a sequence ε→ 0 (not labelled explicitly) such that uε → uµ inL2(Ω)n and ϕ2∇uε ∈L2(Ω)n×n is bounded by the µ-independent upper bound in (4.41). Recall kuµkL2(Ω)n ≤ kfkL2(Ω)n/α from (4.31). Hence, there exists a sequence µ → 0 (not labelled explicitly) such that uµ * u∗ ∈L2(Ω)n converges weakly in L2(Ω)n with ku∗kL2(Ω)n ≤ kfkL2(Ω)n/α andϕ2∇u∗ ∈L2(Ω)n×n bounded by the upper bound in (4.41).
It remains to prove that u∗ =u solves (1.37). Recall thatuµ∈H solves (1.31) and deduce the elementary boundµR
Ω∇uµ: (∇v− ∇uµ)≤ µ4 R
Ω|∇v|2 for any test functionv∈H. Recall that ψ is defined either by (1.35) (Dirichlet) or by (1.33) (Neumann), thatψ(v) =||Dv||L1(Ω)n×n for v∈H, and that α|| • ||L2(Ω)n+ψ(•) is sequentially weakly lower semi-continuous. This and the above weak convergence lead in the limit as µ→0 in (1.31) to
Z
Ω
f ·(v−u∗)≤α Z
Ω
u∗·(v−u∗) +ψ(v)−ψ(u∗) forv∈H. (4.42) Lemma 4.2 below shows thatu∗ =u and the uniqueness of the weak limit also implies uµ* u
inL2(Ω)n not only forsomesequence µ→0 but for any.
Lemma 4.2 Any function u∗ ∈L2(Ω)with (4.42) is equal to the solution u to (1.37).
Proof. In the Dirichlet case, suppose v ∈ L2(Ω)n satisfies ψ(v) < ∞ with ψ defined by (1.35). A modification of Lemma A.1 for BD (rather than BV) functions shows that v can be approximated by a sequence vm ∈ H01(Ω)n such that vm → v in L2 and ψ(vm) → ψ(v) as m→ ∞.
In the Neumann case, suppose v ∈ L2(Ω)n satisfies ψ(v) < ∞ with ψ defined by (1.33). The existence of a sequencevm∈H1(Ω)nsuch thatvm→vinL2(Ω)nandψ(vm)→ψ(v) asm→ ∞ follows from standard regularization [11].
Take the admissible test function v = vm ∈ H in (4.42) for m and then pass to the limit as m → ∞ to deduce that (1.37) holds for the test function v ∈ L2(Ω)n. The uniqueness of the solutions to (1.37) concludes the proof. Note that the assumption ∂Ω Lipschitz applies only
here in each of the two cases.
Remark 4.3 The above passage to the limit as µ→0 and uµ* u in L2(Ω)n also shows
µ→0limµ Z
Ω
|∇uµ|2 = 0 and lim
µ→0ψ(uµ) =ψ(u).
(The proof compares the limit of (4.30) (written in terms ofuµ replacingu therein) with (4.36) and utilizes the sequential weak lower semi-continuity of ψ and ψ(uµ) =kDuµkL1(Ω)n×n.)
4.4 Proof of Theorem 1.8
In the first part of the proof of the characterization, suppose u solves (1.37). Theorem 1.7 asserts that Du ∈L2loc(Ω)n×n∩L1(Ω)n×n and that u is the weak limit in L2(Ω)n, uµ * u as µ → 0, of the solution uµ ∈ H to the viscous problem (1.31). For any µ > 0, the solution uµ ∈H∩Hloc1 (Ω)n satisfies the bound (1.38) and is itself the weak limit in H as δ →0 of the solution uµδ ∈H to the regularized problem
Z
is symmetric and satisfies|σµδ| ≤1 a.e. The pointwise boundedness leads to a sequence δ →0 (not labelled explicitly) such that σµδ *∗ σµ weak star in L∞(Ω)n×n with |σµ| ≤ 1 a.e. The passage to the limit in (4.44) leads to
Z The same arguments apply in the passage to the limit asµ→0 and the weak convergenceuµ* u inL2(Ω)ntowards the solution u to (1.37) and some weak star limitσ ofσµ inL∞(Ω)n×nwith no information on the boundary for test functions in H=H01(Ω)n from (4.47).
The traces in BD are considered in [2] and the normal traces in H(div,Ω) are well known [17].
Forσ ∈H(div,Ω)n∩L∞(Ω)n×nsym, the normal traceσν ∈H−1/2(∂Ω) belongs toL∞(∂Ω)n, as we now show. Since this result can be of independent interest, we state it as a separate lemma.
Lemma 4.4 (traces in BD) If σ ∈ H(div,Ω)n∩L∞(Ω)n×nsym satisfies |σ| ≤ 1 a.e. in Ω, then the normal trace σν ∈H−1/2(∂Ω)n satisfies σν ∈L∞(∂Ω)n with|σν| ≤1 a.e. on the boundary
∂Ωand the integration by parts formula Z
Proof of the lemma. The first step of the proof recalls the tracevb ∈L1(∂Ω)n of a function v∈L2(Ω)n withDv ∈L1(Ω)n×nsym [2] that satisfies the integration by parts formula
Z
Ω
(v·div Φ + Φ :Dv) = Z
∂Ω
vb·Φν for all Φ∈C1(Rn)n×nsym. (4.48) Since vb is unique, we simplify the notation and writev rather thanvb in the Lebesgue integral over the boundary ∂Ω.
Thesecond stepdesigns a sequence of smooth approximations Φj ∈Cc∞(Rn)n×nsym with (i)|Φj| ≤1 in Ω for allj∈N, (ii) its strong convergence Φj →σinH(div,Ω)n, (iii) its weak star convergence Φj *∗ σ inL∞(Ω)n×nsym, and (iv) Φj|∂Ω *∗ σν (weak star) in L∞(∂Ω)n×nsym, asj→ ∞.
It is a standard argument in the approximation of Sobolev functions by mollification in the Lipschitz domain Ω [12, p. 128-129] to consider a suitable partition of unityPN
i=0ζi= 1. Then, forε >0 sufficiently small, one sets Φε:= (ζ0σ)∗ηε+PN
i=1σi∗ηε, where the functionηεis the mollification kernel, σi(x) := (ζiσ)(x+ετi) for all x ∈ Ω∩ωi, and the translation vector τi is such that B(x, ε) +ετi ∈ Ω for all x ∈ Ω∩ωi and all i∈ {1, . . . , N} with ζi supported in ωi. Since ζi is scalar-valued, we have div(ζiσ) =σ∇ζi+ζi(divσ). The arguments in [12] then show that Φε ∈ Cc∞(Rn)n×n with strong convergence to σ in L2(Ω)n×n and strong convergence of divergence div Φε →divσinL2(Ω)n, i.e., strong convergence inH(div,Ω)n. Moreover, Φε takes symmetric values and max|Φε| ≤1 +Cεfor some constant C, that depends only on the smooth and fixed functionsζ0, . . . , ζN in the partition of unity. The pointwise boundedness implies the weak star convergence of a subsequence Φεj.
The third stepsubstitutes Φ = Φεj for the sequence Φεj →σ inH(div,Ω)n from step two for a fixed v∈L2(Ω)n withDv ∈L1(Ω)n×nsym in (4.48). The weak star convergence in L∞(Ω)n×nsym and inL∞(∂Ω)n×nsym leads to the asserted integration by parts formula in the passage to the limit as j→ ∞.
The fourth step concludes the proof: Given any σ := σ(x) ∈ Rn×nsym and ν = ν(x) ∈ Rn with
|σ| ≤ 1 = |ν| (at almost every point x ∈ ∂Ω), the linear functional ϕ7→ ϕ·σν = σ : (ϕν) has a unique Riesz representation z ∈Rn in the Hilbert space (Rn,((•,•))) with respect to the scalar product
((a, b)) := (aν) : (bν) fora, b∈Rn. (4.49) Then ((z, ϕ)) = σ : (ϕν) for all ϕ ∈ Rn implies (zν)ν = σν and |zν|2 = ((z, z)) = σ : (zν)≤ |σ| |zν| ≤ |zν|, whence |zν| ≤ 1. This pointwise definition ofσ(x)7→ z(x) at a.e. x∈∂Ω specifies a measurable function z∈L∞(∂Ω)n. Let us resume the proof of Theorem 1.8. Lemma 4.4 applies to the solutionv=uand its trace ub ∈L1(∂Ω). Substitute divσ =αu−f from (4.46) in the resulting identity and compare with (4.36). This proves
ψ(u) = Z
Ω
σ:Du−β Z
∂Ω
(zν) : (ubν) (4.50) for β = 1 in the Dirichlet case and β = 0 in the Neumann case. (The integration by parts formula (4.47) shows that σν = 0 in the sense of traces in H−1/2(∂Ω)n if H = H1(Ω)n, but traces are involved in the Dirichlet case displayed forβ= 1.) The appendix reports on the traces of BV and BD functions and the weak definition of ψ for the Dirichlet (β = 1) and Neumann (β = 0) case. Lemma A.2 clarifies foru∈L1(Ω)nwith Du∈L1(Ω)n×nthat
ψ(u) = Z
Ω
|Du|+β Z
∂Ω
|ubν|. (4.51)
The equality of (4.50)-(4.51) implies that σ :Du=|Du|a.e. in Ω and, if β = 1, that (zν) : (ubν) =|ubν|a.e. on∂Ω. Since|σ| ≤1 a.e. in Ω and|zν| ≤1 a.e. in∂Ω, the preceding identities imply that equality holds in the Cauchy–Schwarz inequality (pointwise a.e. in Ω).
Consequently, σ and zν are (pointwise a.e. in Ω) a non-negative multiple of Du and ubν, respectively. This leads to the final formulas and concludes the proof of (1.40).
The second part of the proof establishes the converse implication: Suppose u∈L2(Ω)nsatisfies Du∈L1(Ω)n×nand (1.40). Lemma 4.4 applies tov=uand the arising integrals therein satisfy (from (1.40)) that
σ:Du=|Du| a.e. in Ω and σb: (ubν) =−|ubν| a.e. on∂Ω (the second identity holds for β= 1 only). It follows (for β= 0 and β = 1 as above) that
− Z
Ω
u·divσ= Z
Ω
|Du|+β Z
∂Ω
|ubν|.
The combination of this with divσ=αu−f ∈L2(Ω)nand (4.51) proves thatu satisfies (4.36).
Given any v ∈H, an integration by parts with divσ = αu−f ∈L2(Ω)n does not involve any boundary terms (σν = 0 if β = 0 and H =H1(Ω)n, while v ∈ H01(Ω)n for β = 1). This and
|σ| ≤1 a.e and ψ(v) =kDvkL1(Ω)n×n forv∈H verify Z
Ω
(f −αu)·v= Z
Ω
σ :Dv≤ψ(v). (4.52)
The comparison of (4.36) and (4.52) proves thatu∗ :=usatisfies (4.42). Thus Lemma 4.2 shows thatu solves (1.37) and concludes the proof of Theorem 1.8.