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

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688 V.A. Solonnikov

the equation



+ κU(x)+p

=0,x∈G= ∂F, (1.2)

where U(x)=



|x−y|

In what follows we assume that F is a given bounded domain with a smooth

boundary deﬁned by (1.2). The barycenter of F is located on the axis of rotation,

moreover, we assume that



dx =0,i=1, 2, 3.

For simplicity we assume that F is rotationally symmetric with respect to

the x

-axis. The case of non-symmetric F is considered below in Sec. 4.

The functions (1.1) represent a solution of the free boundary problem gov-

erning the evolution of an isolated liquid mass bounded only by a free surface:

+(v ·∇)v − ν∇

v + ∇p =0,

∇·v(x, t)=0,x∈ Ω

,t>0,

T (v,p)n = κU(x, t)n(x), (1.3)

= v ·n,x∈ Γ

= ∂Ω

v(x, 0) = v

(x),x∈ Ω

where Ω

is a domain occupied by the liquid, unknown for t>0 and given for t =0,

is the boundary of Ω

, n is the exterior normal to Γ

, T (v,p)=−pI + νS(v)is

the stress tensor, S(v)=



∂v

∂x

∂v



j,k=1,2,3

is the doubled rate-of-strain tensor,

n is the exterior normal to Γ

, V

is the velocity of evolution of Γ

in the normal

direction,

U(x, t)=



|x −y|

is the Newtonian potential. The solution of (1.3) obeys the conservation laws

|Ω

| = |Ω



v(x, t)dx =



(x)dx,



v(x, t) · η

(x)dx =



(x) ·η

(x)dx, i =1, 2, 3,

where η

(x)=e

× x, e

=(δ

)

j=1,2,3

It is customary to consider the free boundary problem for the perturbations

of V and P written in the coordinate system rotating about the x

-axis with the

same angular velocity ω. It has the form

+(w ·∇)w +2ω(e

×w) − ν∇

w + ∇s =0,

∇·w(y, t)=0,y∈ Ω



,t>0,

T (w,s)n



+ κU



(y,t)+p



, (1.4)

Inversion of the Lagrange Theorem 689



= w · n



,y∈ Γ



w(y, 0) = v

(y) − V (y) ≡ w

(y),y∈ Ω

where Ω



= Z(ωt)Ω

Z(θ)=

⎛

⎝

cos θ −sin θ 0

sin θ cos θ 0

001

⎞

⎠

is a small perturbation of V ,Ω

is a given domain close to F.

To the solution (1.2) of (1.3) corresponds the zero solution of (1.4).

We assume that

|Ω

| = |F|,



(x)dx =



V (x)dx =0,



dx =0,



(x) ·η

(x)dx =



V (x) · η

(x)dx, i =1, 2, 3, (1.5)

which implies

|Ω



| = |F|,





dx =0,i=1, 2, 3,





w(x, t)dx =0,





w(x, t) · η

(x)dx + ω





(x) ·η

(x)dx = ω



(x) ·η

(x)dx, i =1, 2, 3.

(1.6)

If Γ



is close to G, then it can be given by the relation

x = y + N (y)ρ(y, t),y∈G, (1.7)

where N is the exterior normal to G and ρ is a small function. The conditions

(1.5) are equivalent to



ϕ(y,ρ)dS =0,



(y, ρ)dS =0,i=1, 2, 3, (1.8)

where

ϕ(y,ρ)=ρ −

H(y)+

K(y),

(y,ρ)=ϕ(y,ρ)y

+ N

(y)



−

H(y)+

K(y)



H and K are the doubled mean curvature and the Gaussian curvature of G,re-

spectively. The kinematic boundary condition V

= w · n canbewritteninthe

form

(y,t)=

w(x, t) ·n(x)

N(y) · n(x)

,x= y + N (y)ρ(y, t). (1.9)

690 V.A. Solonnikov

Linearization of (1.4), (1.6), (1.8) with respect to w,q,ρ leads to

+2ω(e

×v) − ν∇

v + ∇p =0,

∇·v(x, t)=0,x∈F,t>0,

T (v,p)N + NB

ρ =0, (1.10)

= N(x) · v(y, t) x ∈G= ∂F,

v(x, 0) = v

(x),x∈F,ρ(x, 0) = ρ

(x),x∈G,



ρ(y, t)dS =0,



ρ(y, t)y

dS =0, (1.11)



v(y, t)dy =0,



v(y, t) · η

(y)dy + ω



ρ(y, t)η

(y) · η

(y)dS =0,

i =1, 2, 3, (1.12)

where v(x, t)=(v

), p(x, t), x ∈F,andρ(x, t), x ∈G, are unknown func-

tions,

ρ(x, t)=b(x)ρ −κ



ρ(y, t)dS

|x −y|

b(x)=−ω



·N(x) − κ

∂U(x)

∂N

We assume that b(x) ≥ b

> 0.

Both conditions (1.11) and (1.12) hold for arbitrary t>0, provided they are

satisﬁed for the initial data v

and ρ

The behavior of the solutions of (1.10)–(1.12) for large t is determined by

the spectrum of the corresponding stationary operator. Therefore we consider the

spectral problem

λv +2ω(e

×v) − ν∇

v + ∇p =0,

∇·v(x, t)=0,x∈F,t>0, (1.13)

T (v,p)N + NB

ρ =0,

λρ = N (x) · v(y),x∈G,

in the space of complex-valued functions deﬁned by the conditions (1.11) and

(1.12), i.e.,



ρ(y)dS =0,



ρ(y)y

dS =0, (1.14)



v(y)dy =0,



v(y) · η

(y)dy + ω



ρ(y)η

(y) · η

(y)dS =0,

i =1, 2, 3, (1.15)

Inversion of the Lagrange Theorem 691

Our objective is to prove the following theorem.

Theorem 1. Let

Bρ = B

ρ +









ρ|y



dS. (1.16)

S =



−x

)dx =



−x

)dx > 0(1.17)

and if the quadratic form



ρBρdS (1.18)



b(x)ρ

dS − κ



ρ(x)ρ(y)dS

|x −y|

η



(F)





ρ(x)|x





takes negative values for some ρ(y) satisfying (1.14), then the problem (1.10) has

a ﬁnite number of the points of spectrum with the positive real part.

The quadratic form (1.18) is the second variation of the energy functional.

Therefore the statements similar to that of Theorem 1 are referred to as the in-

version of the Lagrange theorem. On the basis of Theorem 1 the instability of the

zero solution of a nonlinear problem (1.10)–(1.12) can be proved [1].

When the form (1.18) is positive deﬁnite in the space (1.14), then the uni-

formly rotating liquid is stable [2].

The results of the present paper are not quite new. Theorem 1 is proved

in [3, 4]; the proof given here is more complete and simple. As in [3, 4], it is

based on the theorem on invariant subspaces of dissipative operators in the Hilbert

space with indeﬁnite metrics [5]. We follow the ideas of the paper [6] where this

theorem is applied to the problem of motion of a top with cavities ﬁlled with

a viscous incompressible liquid. Another proof of existence of solutions of the

evolution problem (1.10)–(1.12) with the exponential growth for large t>0based

on the construction of the appropriate Lyapunov function is given in [7]. The

method of construction of such function is proposed in [8].

2. Auxiliary propositions

We start with some deﬁnitions and auxiliary relations. We introduce the following

spaces:

J ⊂ L

(F): the space of all the divergence free (in a weak sense) vector ﬁelds

from L

(F),



J: the subspace of vector ﬁelds u ∈ J such that



u(x)dx =0,



u(x) · η

(x)dx =0, (2.1)

692 V.A. Solonnikov

⊥

: the subspace of vector ﬁelds u ∈ J such that



u(x) · η(x)dx =0,

where η(x)=a + b × x is an arbitrary vector of rigid motion,

H: the subspace of functions from L

(G) satisfying the orthogonality conditions

(1.14),

: the subspace of functions from L

(G) satisfying the orthogonality conditions

(1.14) and



ρ(y)y

dS =0,α=1, 2. (2.2)

We set (f,g)=



f(x)g(x)dS and we denote by P



P and P

⊥

orthogonal

(in L

(F)) projections on J,



J and J

⊥

, respectively. We also introduce orthogonal

in L

(G) projections P and P

on H and H

Proposition 1. Arbitrary vector ﬁeld of rigid motion η(x)=a + b × x,wherea

and b are constant vectors, satisﬁes the equation

(η · N)=−ω

η(x) · x



,x∈G. (2.3)

Proof. We take arbitrary small smooth function r(x) and consider the integral

I(r)=





+ κU(x)+p



η(x) · ndS,

where U(x)=



|x −y|

−1

dy and Ω is a domain whose boundary Γ is given by the

equation

x = y + N (y)r(y),y∈G.

It can be shown that only the term containing ω

is diﬀerent from zero; indeed,

we have



η · ndS =



∇·η(x)dx =0and



U(x)n(x)dS =



∇U(x)dx =



y − x

|y − x|

dxdy =0,



U(x)η

(x) ·n(x)dS =



∇U(x) · η

(x)dx



y − x

|y −x|

· η

(x −y)dxdy +



y − x

|y −x|

· η

(y)dxdy

= −



∇U(y) ·η

(y)dy,

from which we can conclude that



U(x)η

(x) ·n(x)dS =0.

Hence

I[r]=





η(x) · n(x)dS = ω



η(x) · x



dx (2.4)

Inversion of the Lagrange Theorem 693

and I[0] = 0. Now we compute the ﬁrst variation of both parts of this equation

with respect to r, making use of the formula

δU = κ

∂U(x)

∂N

+ κ



ρ(y, t)dS

|x −y|

(the proof can be found in [9]). In view of (1.2), we have

−



rη(x) · N(x)dS = ω



r(x)η(x) ·x



dS,

i.e.,



r(x)B

(η(x) · N (x))dS = −ω



r(x)η(x) · x



dS.

Since r(x) is arbitrary, this equation is equivalent to (2.3). The proposition is

proved.

From





η(x) · N(x)dS =2



η(x) · x



dx =0

it follows that also

B(η · N )=−ω

η(x) · x



. (2.5)

Direct computations show that



(x) ·N(x)x

dS = S,



(x) ·N(x)x

dS = −S,



(x) ·N(x)x

dS =



(x) · N(x)x

dS =0. (2.6)

Indeed,



(x) ·N(x)x

dS =



(x) ·∇(x

)dx



−e

)(e

+ e

)dx =



−x

)dx,

andotherequationsareveriﬁedinthesameway.

As a consequence, we obtain



· NB(η

· N)dS = ω



· NdS = ω

which shows that (1.17) is necessary for the positivity of the quadratic form (1.18).

Making use of (2.6), we can easily prove

Proposition 2. An arbitrary ρ ∈ L

(G) can be represented in the form

ρ(x)=ρ

(x)+ρ

(x)(2.7)

694 V.A. Solonnikov

where

(x)=S

−1

(η

(x) ·N(x)I

(ρ) − η

(x) ·N(x)I

(ρ)),

(ρ)=



ρ(x)x

dS, α =1, 2,

and ρ

satisﬁes the orthogonality conditions (2.2).Ifρ ∈ H,thenρ

∈ H

.If

ρ = η

·N, β =1, 2,thenρ

=0.

The equation (2.7) deﬁnes a non-orthogonal projection Q on the space (2.2):

= Qρ = ρ −S

−1



(x) · N(x)I

(ρ) −η

(x) · N(x)I

(ρ)



In view of (2.5), we have (Bρ

,ρ

)=0, which implies

(



Bρ,ρ)=(Bρ,ρ)=(



Bρ

,ρ

)+(



Bρ

,ρ



ρdS =0 (



B is deﬁned in (3.4)). Since

(Bρ

,ρ

)=ω

−1

α=1

(ρ) ≥ 0,

we have

(



BQρ,Qρ) ≤ (



Bρ,ρ). (2.8)

3. On the problem (1.13)–(1.15)

We pass to the analysis of the spectral problem (1.13)–(1.15). By (1.15),

v(x)=v

⊥

(x)+

i=1

(ρ)η

(x),

where v

⊥

= P

⊥

v and

(ρ)=−

η



(F)



ρ(x)η

(x) · η

(x)dS.

We introduce new velocity and pressure according to the formulas

u = v − d

(ρ)η

(x),q= p − ω|x



(ρ)+

|G|



BρdS

and convert (1.13)–(1.15) to

λu +2ω(e

× u) − ν∇

u + ∇q =

ωη

(x)

η



(F)



u · N|x



dS,

∇·u(x)=0,x∈F,

T (v,q)N + N



Bρ =0, (3.1)

λρ = N (x) · u(x) x ∈G,

Inversion of the Lagrange Theorem 695



ρ(y)dS =0,



ρ(y)y

dS =0, (3.2)



u(y)dy =0,



u(y) · η

(y)dy =0,



u(y,t) ·η

(y)dy + ω



ρ(y, t)η

(y) · η

(y)dS =0,j=1, 2, (3.3)

where



Bρ = Bρ −

|G|



BρdS. (3.4)

Since



×u) ·η

(x)dx =



− e

) ·η

(x)dx =



u · N|x



dS,

one can show by a simple calculation that (3.1) is equivalent to

λu +2ω



P (e

×u) −ν∇

u + ∇r =0,

∇·u =0,x∈F, (3.5)

S(u)N − N(N · S(u)N)|

=0,

λρ = N (x) · u(x) x ∈G,

where r is a solution of the Dirichlet problem

∇

r(x)=0,x∈F,r(x)=νN · S(u)N +



Bρ, x ∈G. (3.6)

The pressure is excluded, and we can write (3.5), (3.6), (3.2), (3.3) in the

following abstract form:

λU = AU, (3.7)

where U =(u,ρ)

, A =(A

)

i,j=1,2

u = ν∇

u −∇r

− 2ω



P (e

×u),A

ρ = −∇r

u = u · N ,A

ρ =0,

∇

=0, ∇

=0,x∈F,

= νN · S(u)N,r



Bρ, x ∈G.

As the domain of A, D(A), we take the set U =(u,ρ)

with u ∈ W

(F)

and ρ ∈ W

1/2

(G), satisfying (3.2), (3.3) and the boundary condition

S(u)N − N(N · S(u)N )|

=0.

We also consider a modiﬁed problem

λu +2ω



P (e

×u) −ν∇

u + ∇r =0,

∇·u =0,x∈F, (3.8)

S(u)N − N(N · S(u)N)|

=0,

λρ = N (x) · u(x) − Q

(N · u),x∈G,

696 V.A. Solonnikov

where Q

is the orthogonal projection on the ﬁnite-dimensional space Ker |



that consists of the elements of H satisfying the equation



Bρ = 0. This problem

canbewrittenintheformsimilarto(3.7):

λU = A



where A



is the 2 × 2 matrix operator with



u = A

u,A



ρ = A

ρ,



u = u · N − Q

(u · N),A



ρ =0.

The domain of A



consists of all the elements of D(A) satisfying the additional

orthogonality condition



ρ(x)ϕ(x)dS =0, ∀ϕ ∈ Ker|



B. (3.9)

The following proposition is proved by a direct calculation (see [3]).

Proposition 3. If U =(u,ρ)

∈ D(A



), then (f,g)

= A



U satisﬁes the conditions



gdS =0,



gϕdS =0,



dS =0,i=1, 2, 3,



f(x)dx =0,



f(x) · η

(x)dx =0, (3.10)



f (x) · η

(x)dx + ω



g(x)η

(x) ·η

(x)dS =0,α=1, 2,

where ϕ is an arbitrary element of Ker |



If f and g satisfy (3.10) and λ =0, ±iω, then the solution u ∈ W

(F),

ρ ∈ W

1/2

(G) satisﬁes (3.2), (3.3), (3.9).

The next proposition characterizes the spectrum of A



Proposition 4. The spectrum of A



consists of a countable number of eigenvalues

with the accumulation points λ = ∞ and λ =0. There are no eigenvalues in the

domain Reλ  1 of the complex plane λ and on the imaginary axis.

Proof. We transform the problem (3.1)–(3.3) once more. We use the relations

2ω



× u) · η

dx = ω



u · Nη

·η

dS −ω



u ·η

= ω



u · Nη

·η

dS −ωS

(ρ),

2ω



× u) · η

dx = ω



u · Nη

·η

dS + ω



u ·η

= ω



u · Nη

·η

dS + ωS

(ρ),

Inversion of the Lagrange Theorem 697

where S



+ x

)dx =



+ x

)dx. There relations enable us to write

the ﬁrst equation in (3.5) in the form

−ν∇

⊥

+ ∇r = −λu

⊥

−2ωP

⊥

× u)+ω



(ρ) − η

(ρ)



. (3.11)

Following [10] and [11], Ch. 8, we introduce the operator T

that makes

correspond the solution w

⊥

∈ W

(F) ∩ J

⊥

of the problem

−ν∇

⊥

+ ∇q = f (x),x∈F,

T (w

⊥

,q)N |

to the vector ﬁeld f ∈ J

⊥

, and the operator T

deﬁned by T

ψ = v

⊥

where

⊥

∈ W

(F) ∩ J

⊥

is a solution of the problem

−ν∇

⊥

+ ∇p = l

(ψ)+

i=1

(ψ)η

(x),x∈F, (3.12)

T (v

⊥

,p)N |

= ψ(x)N(x)

with

(ψ)=−

|F|



ψ(x)N(x)dS, l

(ψ)=−

η



(F)



ψη

·NdS,

and ψ is an element of W

1/2

(G) satisfying the condition



ψdS = 0. The operator

canbeextendedtoW

−1/2

(G), and then its range is contained in W

(F) ∩J

⊥

and v

⊥

= T

ψ is a weak solution of (3.12) satisfying the integral identity



S(v

⊥

):S(ϕ)dx =

i=1



(ψ)η

(x) ·ϕ(x)dx



ψ(x)N(x) · ϕ(x)dS, ∀ϕ ∈ W

(F) ∩ J

⊥

(we note that the necessary compatibility conditions

(ψ) ·



η(x)dx +

i=1

(ψ)



(x) ·η(x)dx +



ψ(x)N(x) · η(x)dS =0,

∀η(x)=a + b × x, are satisﬁed).

It is clear that

T

f

(F)

≤ cf 

(F)

T

ψ

(F)

≤ cψ

−1/2

(G)

From (3.11) and from the boundary condition T (u

⊥

,r)N = −



B(ρ)N we can

conclude that

⊥

= −T

(λu

⊥

+2ωP

⊥

×u)) − T

(



Bρ)

= −T

(λu

⊥

+2ωP

⊥

×u)) −

(



B(u · N )). (3.13)