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

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678 S. Shimizu

n−1

j,k=1

− g)

+ g(

−

tjk

)

∂

∂ξ

n−1

j,k=1

− g)∂

+ g(∂

−∂

tjk

)

∂u

∂ξ

n−1

j,k=1

− g

)

∂

g + g

(

−

tjk

)∂

g + g

tjk

(∂

g −∂

)

2gg

∂u

∂ξ

Since

g

−g

1−1/q

(Γ)

≤ C



∇



α

−1/q

(Γ)





∇



+ u

+ v) dτ

(Ω)

+∇



α

1−1/q

(Γ)





∇



+ u

+ v) dτ 

(Ω)



≤ CT

1/p



(2M + R)M,

g

−g

∞

((0,T ),W

1−1/q

(Γ))

≤ CT

1/p



(2M + R)M,



tjk

−



∞

((0,T ),W

1−1/q

(Γ))

≤ CT

1/p



(2M + R)M,

g,

tjk



∞

((0,T ),W

1−1/q

(Γ))

≤ CM,

and for example

(g

− g)

∇



+ u

+ v)

((0,T ),W

−1/q

(Γ))

≤ Cg

−g

∞

((0,T ),W

1−1/q

(Γ))

×



∞

((0,T ),W

1−1/q

(Γ))

∇



+ u

+ v)

((0,T ),L

(Ω))

≤ CT

1/p



(2M + R)

we have

(Δ

− Δ

Γ(t)

)(u

+ u

+ v)

((0,T ),W

−1/q

(Γ))

≤ CT

1/p



(2M + R)

Therefore we obtain

(m −Δ

)

−1

ν ·(Δ

−Δ

Γ(t)

)(u

+ u

+ v)

((0,T ),W

2−1/q

(Γ))

≤ Cν

1−1/q

(Γ)

(Δ

− Δ

Γ(t)

)(u

+ u

+ v)

((0,T ),W

−1/q

(Γ))

≤ CT

1/p



(2M + R)

. (6.34)

Finally we consider the term: (m − Δ

)

−1

ν ·

(t)ξ. By a direct calculation with

Γ(t)



n−1

j=1



∂

tj

−

∂

tj

−

tj

∂

−

tj

∂

tj

∂



Local Solvability of Free Boundary Problems 679

Γ(t)

n−1

j,k=1



tjk

−

tjk



∂

∂ξ

n−1

j,k=1



∂

tjk

−

∂

tjk

−

tjk

∂

−

tjk

∂

tjk

∂



∂α

∂ξ

it holds that

ν ·

(t)ξ =



1+|∇



α|

n−1

j,k=1



∂

tjk

−

tjk

∂



∂

∂ξ

(6.35)

(cf. [24]). Since, from (6.9),

∂

tjk

((0,T ),W

−1/q

(Γ))

≤C∇



α

−1/q

(Γ))

∇



+ u

+ v)

((0,T ),L

(Ω))

≤CT

(2M + R)M,

tjk

∂

((0,T ),W

−1/q

(Γ))

≤CT

(2M + R)M

we have

(m − Δ

)

−1

ν ·

(t)ξ

((0,T ),W

2−1/q

(Γ))

≤ C

∂

tjk

((0,T ),W

−1/q

(Γ))

tjk

∂

((0,T ),W

−1/q

(Γ))

α

3−1/q

(Γ)

≤ C((2M + R)t

. (6.36)

Therefore from (6.27), (6.33), (6.34) and (6.36), we obtain

G(u

+ u

+ v)+ν ·(u

+ u

) ∈ L

((0,T),W

2−1/q

(Γ)),

G(u

+ u

+ v)+ν ·(u

+ u

)

((0,T ),W

2−1/q

(Γ))

≤ C{T

(2M + R)M

+ T

1/p



(2M + R)

}. (6.37)

Step 3 (Estimate of H

and H

). Next we estimate the right-hand side of the

boundary condition H

and H

. From the deﬁnition of H

1,1/2

q,p

(Ω ×R), we have to

extend not only u

and v but also



∇(u

+ u

+ v) dτ to the whole time R.To

do this, we use Lemma 6.3 and the following lemma.

Lemma 6.8 (Lemma 2.4 in [20]). Let 2 <p<∞ and n<q<∞.Set

1,1

q,p

(

Ω ×I)={f ∈ W

1,1

q,∞

(

Ω × I):∂

f ∈ L

(I,W

(

Ω))}.

Let 0 <T ≤ 1. Then, there exist linear operators F

: W

(Ω) →

1,1

q,p

(Ω × R),

: W

2,1

q,p

(Ω × (0, 2)) →

1,1

q,p

(Ω × R), F

: W

2,1

q,p,0

(Ω × (0,T)) →

1,1

q,p

(Ω × R),

680 S. Shimizu

such that

z(ξ,t)=



∇z(ξ,τ) dτ, 0 ≤ t ≤ T,

z =0 for t ∈ [0, 2T ],

F

z

∞

(R,W

(Ω))

≤ CTz

(Ω)

∂

z)

∞

(R,L

(Ω))

≤ Cz

(Ω)

∂

z)

(R,W

(Ω))

≤ Cz

(Ω)

F

z

∞

(R,W

(Ω))

≤ CT

1/p



z

2,1

q,p

(Ω×I

)

for j =2, 3,

∂

z)

∞

(R,L

(Ω))

≤ Cz

2,1

q,p

(Ω×I

)

for j =2, 3,

∂

z)

(R,W

(Ω))

≤ Cz

2,1

q,p

(Ω×I

)

for j =2, 3,

where I

=(0, 2), I

=(0,T),andC is independent of z and T .

We recall that

[[ H

(u)]] = −[[ Π(Π

− Π)(μD(u)ν

)+Π(μD(u)(ν

− ν)) + ΠΠ

(Q(u)ν

)]],

where Q(u) is deﬁned in (1.6). We consider the term H

+Ev) instead of

+ u

+ v) with time integral F

+ F

v,becausewehavetoestimate

the norm of H

1,1/2

q,p

(Ω × R). Since

(

−1

)

= δ

+ B





∇



dτ,...,



∇



j−1

dτ,



∇



j+1

dτ,...,



∇



dτ



(6.38)

where B

is some polynomial such as B

(0,...,0) = 0 (cf. (A.5) Appendix in

[20]), we can write

−1

ν = ν + B





∇udτ



ν,

where (j, k) component of B is given by B

in (6.38). Therefore we obtain

−1

ν|

ν + B(



∇udτ)ν

|ν + B(



∇udτ)ν|

, (6.39)

−ν =

Bν

|(I + B)ν|

−ν

ν · Bν + Bν ·(I + B)ν

|ν||(I + B)ν|(|(I + B)ν| + |ν|)

, (6.40)

∂

Bν

|(I + B)ν|

−

(I + B)ν(I + B)ν) · ∂

Bν

|(I + B)ν|

. (6.41)

By (6.4) and (6.5), we obtain

B(F

+ F

∞

(R,W

(Ω))

≤ CT

1/p



(2M + R), (6.42)

Local Solvability of Free Boundary Problems 681

and by (6.4), (6.40)–(6.42) and (6.18), we obtain

˜ν

t(u

+Ev)



∞

(R,W

(Ω))

≤ CM, (6.43)

ν

t(u

+Ev)

−ν

∞

(R,W

(Ω))

≤B

∞

(R,W

(Ω))

ν

(Ω)

≤ CT

1/p



(2M + R)M. (6.44)

Here and hereafter ν

and ν denote again the extension of ν

and ν, respectively,

on Γ to Ω. By

∂

B(F

+ F

∞

((0,T ),L

(Ω))

≤ Cu

+ E

+ Ev

∞

(R,W

(Ω))

≤ C(2M + R), (6.45)

∂

B(F

+ F

((0,T ),W

(Ω))

≤ Cu

+ E

+ Ev

(R,W

(Ω))

≤ C(2M + R), (6.46)

and (6.4), (6.16), (As.3), (5.1), (5.3), we have

∂

t(u

+Ev)



∞

(R,L

(Ω))

≤ Cu

+ E

+ Ev

∞

(R,W

(Ω))

≤ C(2M + R),

(6.47)

∂

t(u

+Ev)



(R,W

(Ω))

≤ Cu

+ E

+ Ev

(R,W

(Ω))

≤ C(2M + R).

(6.48)

To estimate the norm of H

1,1/2

q,p

(Ω × R), we use the following lemma.

Lemma 6.9 (Lemma 2.6 in [20]). Let 1 <p<∞, n<q<∞ and 0 <T ≤ 1.

Let

1,1

q,p

(Ω × R) bethesamespaceasinLemma6.8.Iff ∈

1,1

q,p

(Ω × R), g ∈

1,1/2

q,p

(Ω × R) and f vanishes when t ∈ [0, 2T ], then we have

fg

1,1/2

q,p

(Ω×R)

≤

p,q

[f

∞

(R,W

(Ω))

(q−n)/(pq)

f



(1−n/(2q))

∞

(R,L

(Ω))

f



n/(2q)

(R,W

(Ω))

]g

1,1/2

q,p

(Ω×R)

Proposition 2.8 in [20] and Lemma 6.3 yield the following lemma.

Lemma 6.10. Let 1 <p,q<∞.Foru

∈ W

2,1

q,p

(Ω×(0, 2)) and v ∈ W

2,1

q,p

(Ω×(0,T)),

there exists a constant C

p,q

> 0 such that

E



1/2

(R,W

(Ω))

≤ C

p,q

u



2,1

q,p

(Ω×(0,2))

Ev

1/2

(R,W

(Ω))

≤ C

p,q

v

2,1

q,p

(Ω×(0,T ))

By (4.2), it holds that

(Π − Π

)d =(d, ν

)ν

−(d, ν)ν =(d, ν

)(ν

− ν)+(d, ν

−ν)ν. (6.49)

Setting f = ν

t(u

+Ev)

and g = D(u

+ E

+ Ev), and applying Lemma 6.9,

we have from (6.44), (6.47), (6.48) and Lemma 6.10,

D(u

+ E

+ Ev)ν

t(u

+Ev)



1,1/2

q,p

(Ω×R)

≤ C(2M + R)M. (6.50)

682 S. Shimizu

Setting f = ν

t(u

+Ev)

− ν and g =(D(u

+ E

+ Ev)ν

t(u

+Ev)

t(u

+Ev)

), and applying Lemma 6.9, we have from Lemma 6.10,

(D(u

+ E

+ Ev)ν

t(u

+Ev)

,ν

t(u

+Ev)

)

×(ν

t(u

+Ev)

− ν)

1,1/2

q,p

(Ω×R)

≤ C(T



M + T

q−n

)(2M + R)

. (6.51)

Combining (6.49), (6.50) and (6.51) we have

Π(Π − Π

)μD(u

+ E

+ Ev)ν

+Ev



1,1/2

q,p

(Ω×R)

≤ C(T



M + T

q−n

)(2M + R)

. (6.52)

In a similar manner we obtain

ΠμD(u

+ E

+ Ev)(ν

t(u

+Ev)

− ν)

1,1/2

q,p

(Ω×R)

≤ C(T



M + T

q−n

)(2M + R)

, (6.53)

ΠΠ

Q(u

+ E

+ Ev)ν

t(u

+Ev)



1,1/2

q,p

(Ω×R)

≤ C(T



+ T

q−n

)(2M + R)

. (6.54)

Therefore by (6.52), (6.53), and (6.54) we obtain

+ E

+ Ev) ∈ H

1,1/2

q,p

(Ω × R),

H

+ E

+ Ev)

1,1/2

q,p

(Ω×R)

≤ C(T



M + T

q−n

)(2M + R)

. (6.55)

From the deﬁnition of H

1,1/2

q,p

(Ω ×(0,T)), we see that

+ u

+ v) ∈ H

1,1/2

q,p

(Ω × (0,T)),

H

+ u

+ v)

1,1/2

q,p

(Ω×(0,T ))

≤ C(T



M + T

q−n

)(2M + R)

. (6.56)

Finally we consider the term H

. We recall that

[[ H

(u, π)]] = [[ν · S(u, π)(ν − ν

) − ν ·Q(u)ν

]]

+[[ρ]] c



dτ ν ·ν

+[[ρ]] c

ν · (ν

−ν)

− 2σ



∇



udτ



∇

· ν + σm



ν · udτ.

We consider the term H

+ E

+ Ev, π

+¯π

θ) instead of H

+ u

v, π

+ π

+ θ) with time integral F

+ F

v,becausewehavetoestimate

the norm of H

1,1/2

q,p

(Ω × R). Similar to the way we estimated H

+ E

+ Ev),

Local Solvability of Free Boundary Problems 683

we have

ν ·(S(u

+ E

+ Ev,π

+¯π

θ)(ν −ν

t(u

+Ev)

)

1,1/2

q,p

(Ω×R)

≤ C(T



M + T

q−n

)(2M + R)

, (6.57)

ν ·Q(u

+ E

+ Ev)ν

t(u

+Ev)



1,1/2

q,p

(Ω×R)

≤ C(T



+ T

q−n

)(2M + R)

. (6.58)

Let χ(t) be a function in C

∞

(R) such that χ(t)=1for|t|≤1andχ(t)=0

for |t|≥2. Setting f = ν

t(u

+Ev)

− ν and g = χ(t)ξ

ν, and applying Lemma

6.9, we have

(ν

t(u

+Ev)

− ν)χ(t)ξ

ν

1,1/2

q,p

(Ω×R)

≤ C(T

1/p



M + T

q−n

)(2M + R)χ(t)ξ

ν

(R,W

(Ω))

≤ C(T

1/p



M + T

q−n

)(2M + R)M

By the deﬁnition of H

1,1/2

q,p

(Ω × (0,T)), we see that

(ν

t(u

+v)

−ν)ξ

ν

1,1/2

q,p

(Ω×(0,T ))

≤ C(T

1/p



M + T

q−n

)(2M + R)M

. (6.59)

By (6.7) we have

(∇

·ν)∇



udτ

1,1/2

q,p

(Ω×(0,T ))

≤ C(T

1/p



M + T

q−n

)(2M + R)M

. (6.60)

Therefore from (6.57), (6.58), (6.7), (6.59) and (6.60), we obtain

+ u

+ v, π

+ π

+ θ) ∈ H

1,1/2

q,p

(Ω × (0,T)),

H

+ u

+ v, π

+ π

+ θ)

1,1/2

q,p

(Ω×(0,T ))

≤ C(T

1/p



M + T

q−n

)(2M + R)M

(6.61)

Step 4. Combining (6.37), (6.56), (6.61), we have

V 

2,1

q,p

(Ω×(0,T ))

+ Θ

((0,T ),

(Ω))

+ 

Θ

1,1/2

q,p

(Ω×(0,T ))

+ ∂

Y 

((0,T ),W

2−1/q

(Γ))

+ Y 

((0,T ),W

3−1/q

(Γ))

≤ C{T

(2M + R)M

+(T



M + T

q−n

)(2M + R)

}. (6.62)

Here we also use

Div Q(u

+ u

+ v),R(u

+ u

+ v)∇(π

+ π

+ θ)

((0,T ),L

(Ω))

≤ CT



(2M + R)

E(u

+ u

+ v)

((0,T ),W

(Ω))

≤ CT



(2M + R)



E(u

+ u

+ v)

((0,T ),L

(Ω))

C(T



+ T

q−n

)(2M + R)

684 S. Shimizu

(cf. §2 in [20]). For given M deﬁned by (6.2), we choose R>0andT>0such

that the right-hand member of (6.62) is less than R, then by (6.62) we have

V 

2,1

q,p

(Ω×(0,T ))

+ Θ

((0,T ),

(Ω))

+ 

Θ

1,1/2

q,p

(Ω×(0,T ))

(6.63)

+ ∂

Y 

((0,T ),W

2−1/q

(Γ))

+ Y 

((0,T ),W

3−1/q

(Γ))

≤ R.

Moreover from the deﬁnition of E(u

+ u

+ v),

E(u

+ u

+ v), H

+ u

+ v)

and H

+ u

+ v, π

+ π

+ θ), we see that

E(u

+ u

+ v)|

t=0

E(u

+ u

+ v)|

t=0

=0,

+ u

+ v)|

t=0

= H

+ u

+ v, π

+ π

+ θ)|

t=0

=0.

Therefore for the map Φ(v, θ,

θ, η)=(V,Θ,

Θ,Y)itmapsI

R,T

into itself.

Now we show that Φ is a contraction map. To do this, given (v

,θ

,η

) ∈

R,T

(i =1, 2), we set

, Θ

)=Φ(v

,θ

,η

),i=1, 2.

From (6.3), we see that V

− V

,Θ

− Θ

and Y

−Y

satisfy the linear equation

∂

− V

) − Div S(V

−V

, Θ

−Θ

)=DivQ(u

+ u

+ v

)

− Div Q(u

+ u

+ v

)+R(π

+ π

+ θ

) − R(π

+ π

+ θ

)inΩ,t>0,

div (V

−V

)=E(u

+ u

+ v

) −E(u

+ u

+ v

)

=div

E(u

+ u

+ v

) − div

E(u

+ u

+ v

)inΩ,t>0,

∂

−Y

) −ν · (V

−V

)=G(u

+ u

+ v

) − G(u

+ u

+ v

)onΓ,t>0,

[[ ΠD(V

−V

)ν]] = [[ H

+ u

+ v

) − H

+ u

+ v

)]] on Γ,t>0,

[[ ν ·S(V

−V

, Θ

− Θ

)ν]] + σ(m − Δ

)(Y

− Y

)

=[[H

+ u

+ v

,π

+¯π

)

−H

+ u

+ v

,π

+¯π

)]] on Γ,t>0,

[[ V

− V

]] = 0 o n Γ ,t>0,

− V

t=0

=0inΩ, (Y

− Y

t=0

=0onΓ. (6.64)

Applying the same argument as in the proof of (6.63), we can show that

V

− V



2,1

q,p

((0,T )×Ω)

+ Θ

− Θ



((0,T ),

(Ω))

+ 

−



1,1/2

q,p

(Ω×(0,T ))

+ Y

− Y



((0,T ),W

2−1/q

(Γ))

≤C{T

+(T



+ T

q−n

)(2M + R)M

}

v

− v



2,1

q,p

((0,T )×Ω)

+ θ

−θ



((0,T ),

(Ω))

+ 

−



1,1/2

q,p

(Ω×(0,T ))

+ η

− η



((0,T ),W

2−1/q

(Γ))

. (6.65)

Local Solvability of Free Boundary Problems 685

Choosing T>0sosmallthat

C{T

+(T



+ T

q−n

)(2M + R)M

}≤

in (6.65), we see that Φ is a contraction map on I

R,T

. Therefore Φ has the ﬁxed

point (V,Θ,

Θ,Y) which solves (6.3). This completes the proof of Theorem 6.1.

Acknowledgment

The author is grateful to Professor Yoshihiro Shibata for the proof of Lemma 2.3

and Proposition 6.7. She is also grateful to the referee for valuable suggestions

which improved the presentation of this paper.

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Senjo Shimizu

Department of Mathematics

Shizuoka University

Shizuoka 422-8529, Japan

e-mail: ssshimi@ipc.shizuoka.ac.jp

Progress in Nonlinear Diﬀerential Equations

and Their Applications, Vol. 80, 687–704

 2011 Springer Basel AG

Inversion of the Lagrange Theorem

in the Problem of Stability of Rotating

Viscous Incompressible Liquid

V.A. Solonnikov

Dedicated to Professor H. Amann on the occasion of his jubilee

Abstract. The paper contains analysis of the spectrum of a linear problem

arising in the study of the stability of a ﬁnite isolated mass of uniformly ro-

tating viscous incompressible self-gravitating liquid. It is assumed that the

capillary forces on the free boundary of the liquid are not taken into account.

It is proved that when the second variation of the energy functional can take

negative values, then the spectrum of the problem contains ﬁnite number of

points with positive real parts, which means instability of the rotating liquid

in a linear approximation. The proof relies on the theorem on the invari-

ant subspaces of dissipative operators in the Hilbert space with an indeﬁnite

metrics.

Mathematics Subject Classiﬁcation (2000). 76E07, 47B50.

Keywords. Rotating liquid, stability, dissipative operators.

1. Formulation of main result

This paper is devoted to the problem of stability of a ﬁnite mass of a viscous

incompressible liquid of a unit density uniformly rotating about a ﬁxed axis (x

axis). The liquid is subject to the forces of self-gravitation; the surface tension on

the free boundary is not taken into account. The velocity and the pressure of the

rotating liquid are given by

V (x)=ω(e

×x),P(x)=



+ p

, (1.1)

where ω is the angular velocity of rotation, e

=(0, 0, 1),x



=(x

, 0), p

const. The domain F occupied by the liquid (the equilibrium ﬁgure) is deﬁned by