Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift

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8H.Abels

Therefore

Δc

+ θP

(c)

+ θ





(c)|∇c|

dx ≤ C∂E

(c)

(3.6)

uniformly in 0 <θ≤ 1 because of θP

(c)=∂E

(c)+Δc. In order to estimate

θm(φ

(c)), we follow some arguments which also can be found in [8, §4] and [11,

Lemma 5.2]. To this end we multiply −Δc +P

θφ

(c)withc ∈ L

(0)

(Ω) and obtain



|∇c|

dx + θ



(c)cdx=(∂E

(c),c)

. (3.7)

Now we choose ε>0sosmallthatφ

c→a

−∞,φ

c→b

∞. (Note that ε depends only on φ

and m =0∈ (a, b).) Then



(c)cdx=



{c(x)∈(a,a+ε]}

(c)cdx+



{c(x)∈(a+ε,b−ε)}

(c)cdx



{c(x)∈[b−ε,b)}

(c)cdx

≥ ε



{c(x)∈(a,a+ε]∪[b−ε,b)}

|φ

(c)|dx − C(ε, Ω)c

(Ω)

because of |φ

(c)|≤C

on [a + ε, b −ε]. Using (3.7), we conclude



|φ

(c)|dx ≤ C(R, ε)(∂E

(c)

+1)

provided that c

≤ R. Combining this with (3.6), we obtain (3.4).

In order to prove (3.5), we multiply −Δc+θφ

(c)withθ

r−1

|φ

(c)|

r−2



(Ω) and obtain

(r − 1)θ

r−1



|φ

(c)|

r−2



(c)|∇c|

dx + θ



|φ

(c)|

≤ C(Ω,r,R)



∂E

(c)

(Ω)

+ θm(φ

(c)



θφ

(c)

r−1

Hence

θφ

(c)

(Ω)

≤ C(R)(∂E

(c)

(Ω)

+ 1) (3.8)

uniformly in 0 <θ≤ 1 provided that c

≤ R. This implies (3.5) because of

(2.1) and ∂

∂Ω

=0. 

Now we consider the limit θ → 0ofE

.Since

lim

θ→0

(c)=



|∇c|

dx +



[a,b]

(c(x)) dx =: E

for all c ∈ H

(Ω), where I

[a,b]

denotes the indicator function of [a, b], which is

deﬁned as

[a,b]

(s)=



0ifs ∈ [a, b],

+∞ else,

Double Obstacle Limit 9

we expect that ∂E

converges to the subgradient of ∂E

in a suitable sense and

under suitable conditions. We note that

∂I

[a,b]

(c)=

⎧

⎪

⎨

⎪

⎩

0ifc ∈ (a, b),

[0, ∞)ifc = b,

(−∞, 0] if c = a,

∅ else.

Moreover, E

is lower semi-continuous on H = L

(0)

(Ω) and H = H

−1

(0)

(Ω) since

lim inf

k→∞

) < ∞ and c

→

k→∞

c in H implies

∈ dom E

, sup

j∈N

∇c



< ∞,c



j→∞

c in H

(Ω) and a.e.

for some subsequence (c

)

j∈N

. Therefore

(c)=

∇c

≤ lim inf

k→∞

∇c



= lim inf

k→∞

The following corollary will be the essential tool for passing to the limit in

the convective Cahn-Hilliard equation (1.3)–(1.4).

Corollary 3.4. Let c

∈ L

∞

(0,T; L

(0)

(Ω)), 0 <T <∞, k ∈ N

,beabounded

sequence, let θ

> 0 be such that θ

→

k→∞

0, and assume that c

(t) →

k→∞

c(t)

in L

(0)

(Ω) and c

(t) ∈D(∂E

) for almost every t ∈ (0,T).If

∂E

) 

k→∞

in L

(0,T; L

(Ω))

for some μ

∈ L

(0,T; L

(0)

(Ω)) and some 1 <q<∞,thenc



k→∞

c in

(0,T; H

(Ω)), μ

(t) ∈ ∂E

(c(t)) for almost every t ∈ (0,T),whereE

is con-

sidered as a functional on L

(0)

(Ω),and

−Δc + P

f = μ

with f (x, t) ∈ ∂I

[a,b]

(c(x, t)) for almost all (x, t) ∈ Q

as well as ∂

c(t)|

∂Ω

=0for almost every t ∈ (0,T). Moreover, if additionally

∂E

) is bounded in L

(0,T; L

(Ω)),forsome2 ≤ r<∞,then

c

(0,T ;W

(Ω))

≤ sup

k∈N

C(R)



1+∂E

)

(0,T ;L

(Ω))



. (3.10)

Proof. Because of (3.4), there is a subsequence k

→

j→∞

∞

) 

j→∞

f in L

(0,T; L

(Ω)).

Moreover, since c

(t) →

k→∞

c(t)inL

(Ω) for almost every t ∈ (0,T)and(c

)

k∈N

is bounded in L

∞

(0,T; L

(Ω)), c

→

k→∞

c in L

(0,T; L

(Ω)) for all 1 ≤ r<∞

by Lebesgue’s theorem on dominated convergence. Together with the boundedness

of c

in L

(0,T; H

(Ω)) due to (3.4) this implies c

→

k→∞

c in L

(0,T; C

(Ω))

because of

f

∞

≤ Cf

1−

f

cf. [3, Equation (2.15)] for a reference.

10 H. Abels

Therefore for a suitable subsequence c

(t) → c(t)inC

(Ω) as j →∞for

almost every t ∈ (0,T). Therefore

(x, t)) →

j→∞

0=f(x, t)a.e.in{(x, t) ∈ Q

: c(x, t) ∈ (a, b)}.

On the other hand, if c(x, t)=a for some x ∈ Ωandsomet ∈ (0,T) such that

(t) converges strongly in C(Ω), then φ

(x, t)) ≤ 0 for suﬃciently large j and

therefore f(x, t) ≤ 0, i.e., f(x, t) ∈ ∂I

[a,b]

(a), almost everywhere on {c(x, t)=a}.

Bythesameargumentf(x, t) ≥ 0, i.e., f(x, t) ∈ ∂I

[a,b]

(b)almosteverywhereon

{c(x, t)=b}. On the other hand

f = lim

j→∞

) = lim

k→∞

(∂E

)+Δc

)=μ

+Δc

weakly in L

(0,T; L

(Ω)). Hence it only remains to prove μ

(t) ∈ ∂E

(c(t)) for

almost every t ∈ (0,T). But this follows from the fact that



η(t)(μ

(t),c



− c(t))

(0)

dt = lim

k→∞



η(t)(∂E

(t)),c



−c

(t))

(0)

≤ lim

k→∞



η(t)(E



) −E

(t))) dt



η(t)(E



) − E

(c(t))) dt

for all c



∈ dom E

=domE

and η ∈ C

∞

(0,T)withη ≥ 0sinceθ

(s) →

k→∞

0 uniformly in s ∈ [a, b]andc

(t) →

k→∞

c(t)inC(Ω) for almost all t ∈ (0,T).

Finally, (3.10) follows from (3.8), ∂

∂Ω

=0,and(2.1). 

For completeness we give a characterization of ∂E

and D(∂E

), which is

based on the results of Kenmochi et al. [11].

Lemma 3.5. Let E

be as above and let ∂E

be its subgradient with respect to

(0)

(Ω).Then

D(∂E



c ∈ H

(Ω) ∩ L

(0)

(Ω) : c(x) ∈ [a, b] a.e.,∂

∂Ω



(3.11)

and μ

∈ ∂E

(c), c ∈D(∂E

),ifandonlyifc ∈ H

(Ω) ∩L

(0)

(Ω), ∂

∂Ω

=0,and

β(x):=μ

(x)+Δc(x)+μ ∈ ∂I

[a,b]

(c(x)) a.e. in Ω (3.12)

for some μ ∈ R. Moreover, c

+ P

β

≤ C (μ



+ c

Finally, if



is the extension of E

to H

−1

(0)

(Ω),thenw ∈ ∂



c ∈D(∂



(c)) if and only if w = −Δ

,whereμ

∈ H

(Ω) ∩ ∂E

(c).

Proof. First let c ∈D(∂E

)andletμ

∈ ∂E

(c). Then by deﬁnition

(μ



− c)

≤ E



) − E



∈ dom(E

Double Obstacle Limit 11

Hence c is a minimizer of the functional F (c



)=E



) − (μ



)

deﬁned on

(0)

(Ω). Since F (c



) is strictly convex, the minimizer is unique. Therefore [11,

Proposition 5.1, Lemma 5.3] imply that Δ

c ∈ L

(Ω) and

−Δ

c + β = μ

+ μ,

where β(x) ∈ ∂I

[a,b]

(c(x)) for almost all x ∈ Ω, μ ∈ R,andΔ

c

+ P

β

≤

Cμ



. Hence c ∈ H

(Ω), ∂

∂Ω

= 0, and (3.12) holds. This proves one impli-

cation and one set inclusion in (3.11).

To prove the converse implication let c ∈ H

(Ω) ∩ L

(0)

(Ω), ∂

∂Ω

=0,and

let μ

∈ L

(0)

(Ω) satisfy (3.12). Then

(μ

+Δc + μ, c



− c)

≤





[a,b]



) − I

[a,b]

(c)



dx =0

for all c



∈ dom E

since μ

+Δc + μ ∈ ∂I

[a,b]

(c). Moreover, using

(Δc, c



− c)

∇c

−

∇c





∇(c −c



)

we conclude

(μ



− c)

≤ E



) − E



∈ dom E

i.e., μ

∈ ∂E

(c). Furthermore, for every c ∈ H

(Ω) ∩L

(0)

(Ω) with ∂

∂Ω

=0we

have that −Δc ∈ ∂E

(c)since



) − E

(c)=



∇c ·∇(c



− c) dx +



|∇(c



− c)|

dx ≥−



Δc(c



−c) dx

for all c



∈ dom(E

). Therefore (3.11) holds.

In order to prove the last statement, let w ∈ ∂



(c). Then μ

= −Δ

−1

w ∈

(Ω) ∩ ∂E

(c)since

(μ



−c)

=(w, c



− c)

−1

(0)

≤





) −



(c)=E



) − E

(c)

for all c



∈ dom E

=dom



.Conversely,ifμ

∈ H

(Ω) ∩ ∂E

(c), then w :=

−Δ

∈ H

−1

(0)

(Ω) ∈ ∂



3.2. Convective Cahn-Hilliard equation

In this section we consider

∂

c + v ·∇c =Δμ in Ω ×(0, ∞), (3.13)

μ = φ(c) − Δc in Ω × (0, ∞), (3.14)

∂

∂Ω

= ∂

μ|

∂Ω

=0 on∂Ω × (0, ∞), (3.15)

t=0

= c

in Ω (3.16)

12 H. Abels

for given c

with E

free

) < ∞ and v ∈ L

∞

(0, ∞; L

(Ω)) ∩L

(0, ∞; H

(Ω)). Here

φ =Φ



where Φ is as in (1.12)–(1.13) and E

free

(1.12) yields the decomposition

free

(c)=E

c

(Ω)

where E

(c), respectively

to its extension to H

−1

(0)

(Ω), which is denoted by



(c).

We note that we can consider (3.13)–(3.16) as an evolution equation on

−1

(0)

(Ω):

∂

c(t)+A

(c(t)) + B(v(t))c(t)=0, for t>0, (3.17)

t=0

= c

(3.18)

where A

(c)=∂



(c)and

B(v)c, ϕ

−1

(0)

=(v ·∇c, ϕ)

− θ

(∇c, ∇ϕ)

for all c, ϕ ∈D(B(v)) = H

(0)

(Ω). This means that A

(c)=Δ

(Δc−θP



(c)) due

to Corollary 3.1 and B(v)c = v ·∇c + θ

c,whereΔ

: H

(0)

(Ω) ⊂ H

−1

(0)

(Ω) →

−1

(0)

(Ω) is the Laplace operator with Neumann boundary conditions, which is

considered as an unbounded operator on H

−1

(0)

(Ω). Finally, we note that A

is a

strictly monotone operator since

) −A

),c

− c

)

−1

(0)

=(−Δ(c

−c

)+θφ

) − θφ

),c

− c

)

≥∇(c

− c

)

(3.19)

for all c

∈D(A

From [3] we recall:

Theorem 3.6. Let v ∈ L

(0, ∞; H

(Ω)) ∩ L

∞

(0, ∞; L

(Ω)). Then for every c

∈

(0)

(Ω) with M := E

free

) < ∞ there is a unique solution

c ∈ BC([0, ∞); H

(0)

(Ω))

of (3.13)–(3.16) with ∂

c ∈ L

(0, ∞; H

−1

(0)

(Ω)), μ ∈ L

uloc

([0, ∞); H

(Ω)).This

solution satisﬁes

free

(c(t)) +



|∇μ|

d(x, τ )=E

free

) −



v · μ∇cd(x, τ) (3.20)

for all t ∈ [0, ∞) and

c

∞

(0,∞;H

)

+ ∂

c

(0,∞;H

−1

(0)

)

+ ∇μ

(Q)

≤ C



M + v

(Q)



(3.21)

c

uloc

([0,∞);W

)

+ φ(c)

uloc

([0,∞);L

)

≤ C



M + v

(Q)



(3.22)

Double Obstacle Limit 13

where r =6if d =3and 1 <r<∞ is arbitrary if d =2.HereC, C

are

independent of v, c

. Moreover, for every R, T > 0 the solution

c ∈ Y := L

(0,T; W

(Ω)) ∩ H

(0,T; H

−1

(0)

(Ω))

depends continuously on

,v) ∈ X := H

(Ω) × L

(0,T; L

(Ω)) such that E

free

)+v

(0,∞;H

)

≤ R

with respect to the weak topology on Y and the strong topology on X.

Remark 3.7. For ﬁxed θ (θ = 1) the theorem coincides with [3, Theorem 6], where

we note that (3.21) is only based on the energy estimate and (3.22) is based on

(3.3). Since the constant in (3.3) can be chosen independent of 0 <θ≤ 1, cf.

(3.5), it is easy to observe from the proof of [3, Theorem 6] that the inequalities

(3.21)–(3.22) hold uniformly with respect to 0 <θ≤ 1 due to Proposition 3.3.

The following improved regularity statement will be important to get higher

regularity of solutions to the Navier-Stokes/Cahn-Hilliard system.

Lemma 3.8. Let the assumption of Theorem 3.6 be satisﬁed and let (c, μ) be the

corresponding solution of (3.13)–(3.16). Moreover, let 0 ≤ s<

,letω ≡ 1 if

∈D(∂



E) and let ω(t)=



1+t



else.

1. If ∂

v ∈ L

uloc

([0, ∞); H

−s

(Ω)) and r is as in Theorem 3.6,then(c, μ) satisﬁes

ω∂

c ∈ L

∞

(0, ∞; H

−1

(0)

(Ω)) ∩ L

uloc

([0, ∞); H

(Ω)),

ωc ∈ L

∞

(0, ∞; W

(Ω)),ωφ(c) ∈ L

∞

(0, ∞; L

(Ω)),ωμ∈ L

∞

(0, ∞; H

(Ω)).

2. If v ∈ B

∞,uloc

([0, ∞); H

−s

(Ω)) for some α ∈ (0, 1),then

ωc ∈ C

([0, ∞); H

−1

(0)

(Ω)) ∩ B

2∞,uloc

([0, ∞); H

(Ω)). (3.23)

Finally, the same statements hold true if [0, ∞) is replaced by [0,T], 0 <T <∞,

Remark 3.9. The lemma coincides with [3, Lemma 3], where θ>0wasﬁxed.

As in Remark 3.7, the estimates in the proof of the latter lemma hold uniformly

with respect to 0 <θ≤ 1 since they are based essentially on (3.21)–(3.22) and

(3.19). Hence Proposition 3.3 implies that all estimates obtained in the proof of

[3, Lemma 3] are independent of θ ∈ (0, 1].

3.3. Limit for the convective Cahn-Hilliard equation

In this section we show that solutions of the convective Cahn-Hilliard equation

(1.3)–(1.4) with φ(c)=θφ

c converge as θ → 0 to solutions of the system

∂

c + v ·∇c =Δμ in Ω × (0, ∞), (3.24)

μ +Δc + θ

c ∈ ∂I

[a,b]

(c)inΩ× (0, ∞), (3.25)

∂

∂Ω

= ∂

μ|

∂Ω

=0 on∂Ω ×(0, ∞), (3.26)

t=0

= c

in Ω (3.27)

14 H. Abels

for given c

with E

free

) < ∞ and v ∈ L

∞

(0, ∞; L

(Ω))∩L

(0, ∞; H

(Ω)), where

free

(c)=



|∇c|

dx +





[a,b]



dx. (3.28)

Again we have a decomposition

free

(c)=E

c

(Ω)

where E



(x) dx = 0. As before,

(3.24)–(3.27) can be written as an abstract evolution equation in H

−1

(0)

(Ω):

∂

c(t)+A

(c(t)) + B(v(t))c(t)  0,t>0, (3.29)

t=0

= c

(3.30)

where A

(c)=∂



(c), cf. Lemma 3.5, is now a multi-valued maximal monotone

operator and B is as before. Moreover, we have as before

−w

− c

)

(0,∞;H

−1

(0)

)

= −(Δ(c

−c

)+f

− f

−c

)

(Q)

≥c

− c



(0,∞;H

(0)

)

(3.31)

where w

(t) ∈ ∂



(t)), j =1, 2, for almost every t ∈ (0, ∞)andf

(x, t) ∈

∂I

[a,b]

(x, t)), j =1, 2, almost everywhere.

More precisely, we show

Theorem 3.10. Let 0 <θ

≤ 1, k ∈ N, be such that θ

→

k→∞

0. Moreover, assume

0,k

∈ H

(0)

(Ω) with c

0,k

(x) ∈ [a, b] almost everywhere for all k ∈ N

and assume

that v

∈ L

(0, ∞; H

(Ω)) ∩ L

∞

(0, ∞; L

(Ω)) such that



k→∞

v in L

(0,T; H

(Ω)),c

0,k

→

k→∞

in H

(0)

(Ω)

for all 0 <T<∞,andlet(c

,μ

) denote the unique solutions of (3.13)–(3.16)

with (v, c

,φ(c)) replaced by (v

0,k

,φ

(c)),whereφ

(c)=Φ



(c)=

.Then



k→∞

c in L

(0,T; W

(Ω)),

∇μ



k→∞

∇μ in L

(Q),

) 

k→∞

f in L

(0,T; L

(Ω))

for all 0 <T <∞,wheref = μ +Δc ∈ ∂I

[a,b]

if d =3, 2 ≤ r<∞ is arbitrary if d =2,and(c, μ) ∈ BC([0, ∞); H

(Ω)) ∩

uloc

([0, ∞); H

(Ω)) istheuniquesolutionof (3.24)–(3.27). Moreover, for every

t>0

free

(c(t)) +



|∇μ|

d(x, τ )=E

free

) −



v · μ∇cd(x, τ). (3.32)

Double Obstacle Limit 15

Proof. First of all, because of Remark 3.7, (3.21)–(3.22) with (c, μ, φ(c),v,c

)re-

placed by (c

,μ

,φ

(c),v

0,k

) hold true with constants independent of θ

. Hence

for a suitable subsequence k

→

j→∞

∞



j→∞

c in L

(0,T; W

(Ω)),

∇μ



j→∞

∇μ in L

(Q),

) 

j→∞

f in L

(0,T; L

(Ω))

for all 0 <T <∞.Moreover,since∂

is bounded in L

(0, ∞; H

−1

(0)

(Ω)) due to

(3.21),

→

j→∞

c in L

(0,T; W

(Ω))

for all 0 <T <∞ due to the lemma of Aubin-Lions. Hence

B(v



j→∞

B(v)c in L

(0, ∞; H

−1

(0)

(Ω))

and (3.24) holds in the sense of distributions.

Moreover, since c

∈ H

(0,T; H

−1

(0)

(Ω)) is bounded and c

converges strongly

in L

(0,T; H

(Ω)) for all 0 <T<∞, c

converges weakly in H

(0,T; H

−1

(0)

(Ω)) →

([0,T]; H

−1

(0)

(Ω)) for every 0 <T <∞, which implies that c|

t=0

= c

lim

j→∞

0,k

holds in H

−1

(0)

(Ω). Hence c ∈ BC

([0, ∞); H

(Ω)) due to Lemma 2.1

and since c

∈ L

∞

(0, ∞; H

(Ω)) is bounded. Furthermore, since c

converges

strongly in L

(0,T; W

(Ω)) for every 0 <T <∞, a suitable subsequence of c

(t)

(again denoted by c

(t)) converges in W

(Ω) → C

(Ω) for almost all t ∈ (0, ∞).

Because of Corollary 3.4, we conclude that

(μ + θ

c)=−Δc + P

f ∈ ∂E

(c)

with f(x, t) ∈ ∂I

[a,b]

(c(x, t)) almost everywhere. Deﬁning m(μ) by the equation

m(μ)+θ

m(c)=m(f), we obtain (3.25). Moreover,

∂



)=−Δ

(−Δc

+ P

))

= −Δ

(μ

+ θ

) 

j→∞

−Δ

(μ + θ

c)=−Δ

(−Δc + P

in L

(0,T; H

−1

(0)

(Ω)) for every 0 <T <∞. Hence −Δ

(−Δc(t)+P

f(t)) ∈

∂



(c(t)) for almost every 0 <t<∞ because of Lemma 3.5 and (3.17) holds.

Finally, the uniqueness and (3.32) is proved in the same way in Theorem 3.6

using (3.31), cf. [3, Proof of Theorem 6]. Since every convergent subsequence con-

verges to some (c, μ) solving (3.24)–(3.27) and the limit is unique, the complete

sequence converges to the unique solution of (3.24)–(3.27). 

Finally, we state the analogous result to Lemma 3.8 for the limit system

(3.24)–(3.27).

Lemma 3.11. Let v ∈ L

(0, ∞; H

(Ω))∩L

∞

(0, ∞,L

(Ω)),c

∈ H

(Ω) with c

(x) ∈

[a, b] almost everywhere and let (c, μ) be the corresponding solution of (3.24)–

(3.27). Moreover, let ω ≡ 1 if c

∈D(∂



) ∩H

(Ω) and let ω(t)=



1+t



else.

16 H. Abels

1. If ∂

v ∈ L

uloc

([0, ∞); H

−s

(Ω)) for some 0 ≤ s<

and r is as in Theo-

rem 3.10,then(c, μ) satisfy

ω∂

c ∈ L

∞

(0, ∞; H

−1

(0)

(Ω)) ∩ L

uloc

([0, ∞); H

(Ω))

ωc ∈ L

∞

(0, ∞; W

(Ω)),ωφ(c) ∈ L

∞

(0, ∞; L

(Ω)),ωμ∈ L

∞

(0, ∞; H

(Ω)).

2. If v ∈ B

∞,uloc

([0, ∞); H

−s

(Ω)) for some 0 ≤ s<

and α ∈ (0, 1),then

ωc ∈ C

([0, ∞); H

−1

(0)

(Ω)) ∩ B

2∞,uloc

([0, ∞); H

(Ω)). (3.33)

Finally, the same statements hold true if [0, ∞) is replaced by [0,T), T<∞,

Proof. Let θ

, k ∈ N, v

= v, c

0,k

= c

if c

∈D(∂



) ∩ H

(Ω). If

∈D(∂E

) ∩ H

(Ω), then let c

0,k

= c

(ε

x), where ε

> 0 is chosen such that

(s)| +



(s) ≤ C on [ε

a, ε

b]andε

→

k→∞

0, where we assume w.l.o.g.

m =0∈ (a, b). Then c

0,k

∈D(∂



)and∂



0,k

)

≤ C(c



+1).

Moreover, let c

,μ

be as in Theorem 3.10. Then, because of Proposition 3.3, the

estimates in the proof of Lemma 3.8 hold uniformly in 0 <θ

≤ 1, cf. Remark 3.9.

Hence c

are bounded in the corresponding spaces and the lemma follows. 

4. Main result: Double obstacle limit for the model H

First of all, we recall the deﬁnition of weak solutions to (1.1)–(1.7) and a theorem

on existence of weak solutions from [3]:

Deﬁnition 4.1. (Weak Solution)

Let 0 <T ≤∞. A triple (v, c,μ) such that

v ∈ BC

(0,T; L

(Ω)) ∩ L

(0,T; V (Ω)),

c ∈ BC

(0,T; H

(Ω)),φ(c) ∈ L

loc

([0,T); L

(Ω)), ∇μ ∈ L

)

is called a weak solution of (1.1)–(1.7) on (0,T)if

−(v, ∂

ψ)

−(v

,ψ|

t=0

)

+(v ·∇v, ψ)

+(ν(c)Dv, Dψ)

=(μ∇c, ψ)

(4.1)

for all ψ ∈ C

∞

(0)

([0,T) × Ω)

with div ψ =0,

−(c, ∂

ϕ)

− (c

,ϕ|

t=0

)

+(v ·∇c, ϕ)

= −(∇μ, ∇ϕ)

(4.2)

(μ, ϕ)

=(φ(c),ϕ)

+(∇c, ∇ϕ)

(4.3)

for all ϕ ∈ C

∞

(0)

([0,T) × Ω), and if the (strong) energy inequality

E(v(t),c(t)) +



,t)

ν(c)|Dv|

d(x, τ)+



,t)

|∇μ|

d(x, τ ) ≤ E(v(t

),c(t

))

(4.4)

holds for almost all 0 ≤ t

<T including t

=0andallt ∈ [t

,T).

Double Obstacle Limit 17

Theorem 4.2. (Global Existence of Weak Solutions, [3, Theorem 1])

For every v

∈ L

(Ω), c

∈ H

(Ω) with c

(x) ∈ [a, b] almost everywhere there is

aweaksolution(v, c,μ) of (1.1)–(1.7) on (0, ∞). Moreover, if d =2,then(4.4)

holds with equality for all 0 ≤ t

≤ t<∞. Finally, every weak solution on (0, ∞)

satisﬁes

∇

c, φ(c) ∈ L

loc

([0, ∞); L

(Ω)),

1+t

c ∈ BUC(0, ∞; W

(Ω)) (4.5)

where r =6if d =3and 1 <r<∞ is arbitrary if d =2and q>3 is independent

of the solution and initial data. If additionally c

∈ H

(Ω) := {c ∈ H

(Ω) :

∂

∂Ω

=0} and −Δc

+ θφ

) ∈ H

(Ω),thenc ∈ BUC(0, ∞; W

(Ω)).

We show that weak solutions of (1.1)–(1.7) converge as θ → 0 (for a suitable

subsequence) to a weak solution of

∂

v + v ·∇v − div(ν(c)Dv)+∇p = μ

∇c in Ω × (0, ∞), (4.6)

div v =0 inΩ× (0, ∞), (4.7)

∂

c + v ·∇c =Δμ in Ω × (0, ∞), (4.8)

μ +Δc + θ

c ∈ ∂I

[a,b]

(c)inΩ× (0, ∞) (4.9)

together with (1.5)–(1.7). The deﬁnition of weak solutions to (4.6)–(4.9), (1.5)–

(1.7) is the same as in Deﬁnition 4.1 just replacing E

free

(c)byE

free

(c)and(4.3)by

(4.9) together with ∂

∂Ω

= 0, assuming c ∈ L

loc

([0, ∞); H

(Ω)) in the deﬁnition

of weak solutions. Here E

free

(c)=E

c

(Ω)

is as in (3.28).

Our main result of this section is the following:

Theorem 4.3. Let d =2, 3, θ

> 0, k ∈ N

such that θ

→

k→∞

0. Moreover, let

,μ

) be weak solutions of (1.1)–(1.7) with initial values (v

0,k

) →

k→∞

) in L

(Ω) × H

(Ω) with c

0,k

(x) ∈ [a, b] for almost all x ∈ Ω and all k ∈ N.

Then there is a subsequence k

, j ∈ N

, k

→

j→∞

∞ such that

, ∇μ

) 

j→∞

(v, ∇μ) in L

(0, ∞; H

(Ω)

×L

(Ω)), (4.10)

) 

∗

j→∞

(v, c) in L

∞

(0, ∞; L

(Ω)

× H

(Ω)), (4.11)

,μ

) 

j→∞

(c, μ) in L

(0,T; W

(Ω) × L

(Ω)) (4.12)

for all 0 <T <∞ with r =6if d =3and 2 ≤ r<∞ arbitrary if d =2

and (v, c, μ) is a weak solution of (4.6)–(4.9), (1.5)–(1.7). Moreover, every weak

solution of (4.6)–(4.9), (1.5)–(1.7) satisﬁes

∇

c, μ ∈ L

uloc

([0, ∞); L

(Ω)),κ(t)c ∈ BUC([0, ∞); W

(Ω))

for some q>dand with κ ≡ 1 if c

∈D(∂



) and κ(t)=t

/(1 + t)

else.

Proof. By the energy estimate v

∈ L

∞

(0, ∞; L

(Ω)) ∩ L

(0, ∞; H

(Ω)), c

∈

∞

(0, ∞; H

(Ω)), ∇μ

∈ L

(Q) are uniformly bounded. Hence there is a sub-

sequence such that (4.10)–(4.11) holds. Moreover, since (3.21)–(3.22) hold uni-

formly in k ∈ N, cf. Remark 3.7, we can extract a subsequence such that (4.12)