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698 V.A. Solonnikov

Finally, we add the equation

(ρ)+η

(ρ)=

ωη



ρ(x)x

dS +

ωη



ρ(x)x

= −

(x)

ωS





Bρη

· NdS +

(x)

ωS





Bρη

·NdS

= −

(x)

λωS





B(u · N )η

· NdS +

(x)

λωS





B(u · N )η

· NdS

to (3.13) and obtain

u = −λT

⊥

u − 2ωT

⊥

× u) −



B(u · N ) ≡L(λ)u, (3.14)

where

ψ = T

ψ +

(x)

ωS



ψη

· NdS −

(x)

ωS



ψη

·NdS.

Since u · N ∈ W

−1/2

(G) for arbitrary u ∈ J, the operator L(λ) is well deﬁned



J and is completely continuous, moreover, L(λ) is a holomorphic function of λ

everywhere except for λ =0andλ = ∞.

We have shown that (3.5), (3.6) imply the equation (3.14) for u ∈



J;nowwe

prove that (3.5), (3.6) follow from (3.14). Indeed, if (3.14) holds, then

⊥

= −T

(λu

⊥

+2ωP

⊥

× u)) −

(



B(u ·N )),



u(x) ·η

(x)dx = −

λω





B(u · N )η

· NdS



(u(x) · N(x) − Q

(u ·N ))x

dS,



u(x) ·η

(x)dx =



(u(x) · N(x) − Q

(u ·N ))x

dS.

From the smoothing properties of T

and T

mentioned above it follows that the

equation (3.11) is satisﬁed with ρ = λ

−1

(I −Q

)u ·N and as a consequence (3.2),

(3.3), (3.8), (3.9) hold.

We are in a position to apply the theorem on the holomorphic operator

function to L(λ). One of the assumptions of this theorem is that (3.14) implies

u = 0 for a certain λ = 0. We verify this assumption for λ in the half-plane Reλ ≥

L  1. To this end, we multiply the ﬁrst equation in (3.8) by u and integrate the

product over F. We integrate by parts and use the boundary conditions, which

leads to

λu

+2ω



¯u

−¯u

)dx+





B(u ·N )

u ·N dS +

S(u

⊥

)

=0, (3.15)

where ·is the norm in L

(F). Moreover, from

λρ =(I − Q

)u ·N

Inversion of the Lagrange Theorem 699

it follows that

λd

(ρ)=d

(u · N)=d

⊥

· N)+

β=1

(ρ)d

(η

· N),α=1, 2.

Hence for large Reλ we have

α=1

(ρ)|≤c

β=1

⊥

·N)| and





B(u · N )

u · NdS



B(u · N )

u ·NdS



B(u

⊥

·N +

α=1

(ρ)η

· N)(

⊥

· N +

α=1

(ρ)η

· N)dS

≤ c



⊥

·N|

dS. (3.16)

From (3.15) and (3.16) we can conclude that

Reλu

S(u

⊥

)

≤ cReλ

−1



⊥

· N|

dS,

and, as a consequence, that u =0andρ =0,ifReλ is suﬃciently large.

According to the theorem on the holomorphic operator function [12], the

spectrum of the pencil I −L(λ) consists of a countable number of eigenvalues with

the accumulation points at zero and inﬁnity. It can be veriﬁed that every λ =0

that does not coincide with these eigenvalues belongs to the resolvent set of the

operator A



.Letu ∈



J be the solution of the equation

u = L(λ)u + h, where h = −



Bg + T

⊥

f ∈



with arbitrary f ∈



J, g ∈ W

3/2

(G) satisfying (3.10). By repeating the above

arguments it is not diﬃcult to show that U =(u,ρ)

with ρ = λ

−1

(u · N + g)is

asolutionof

λU = A



U + F, F =(f ,g)

From the estimate

u≤ch

and from the properties of T

and T

it follows that

u

(F)

+ ρ

1/2

(G)

≤ c



f

(F)

+ g

3/2

(G)



This completes the proof of the ﬁrst statement of the proposition.

Now we pass to the proof of the second statement. Let λ = 0 be an imaginary

eigenvalue of A



. We multiply the ﬁrst equation in (3.8) by u(x) and integrate over

F. Then we integrate by parts and use the boundary conditions. This leads to

λu

(F)

+2ω



¯u

−u

¯u

)dx −





B(ρ)

u · NdS +

S(u

⊥

)

(F)

=0.

(3.17)

700 V.A. Solonnikov

Since





B(ρ)

u · N dS =





B(ρ)(

u · N − Q

(

u · N))dS =



Bρ¯ρdS,

the ﬁrst three terms in (3.17) are imaginary, hence S(u

⊥

)

(F)

=0,u

⊥

(x)=0

and u(x)=

α=1

(ρ)η

(x). The ﬁrst equation in (3.1) reduces to

α=1

+2ω(e

α=1

)+∇q

= η

(x)

η



(F)



α=1

(y) · N (y)|y



dS =0.

Applying the operator rot to this equation one arrives at

2λ

α=1

=2ω(d

− d

i.e., to

λd

(ρ)=ωd

(ρ),λd

(ρ)=−ωd

(ρ). (3.18)

Moreover, taking the relations d

·N ) = 0 and (2.6) into account, we obtain

λd

(ρ)=d

(

α=1

(ρ)η

· N)=−

ωS

(ρ),λd

(ρ)=

ωS

(ρ). (3.19)

From (3.18), (3.19) it follows that d

= d

=0, and, as a consequence, u =0,

q =0,ρ =0.

Finally, if u,ρis a solution of (3.8) with λ = 0, then the same arguments yield

⊥

=0andu =

α=1

(ρ)η

(x). Multiplying the equation (I −Q

)u ·N |

by x

and integrating, we arrive at



u ·Nx

dS =0, i.e., at

α=1

(ρ)



(x) ·N(x)x

dS =0,β=1, 2.

This implies d

= d

= 0, hence ∇q =0,q = q

=const. From the boundary

condition −q

+ B(ρ) = 0 it follows that



B(ρ)=0. In view of (3.9), this implies

ρ = 0. The proposition is proved.

Proof of Theorem 1. We go back to Proposition 2. By (2.8), the quadratic form

(



Br,r) can take negative values for some r ∈ H

. We introduce the space H

- Ker



B and the projection P

on this space. We note that B

= P



bI + K, where b(x) > 0andK is an integral operator with a weakly singular

kernel, hence B

has a ﬁnite number of negative eigenvalues (see [14], Ch. 10). We

denote by H

−

the space spanned by the corresponding eigenfunctions of B

.For

r ∈ H

−

, r =0,wehave(B

r, r) < 0andr ∈ H

- H

−

≡ H

, r =0,satisfy

the inequality (B

r, r) > 0. Hence H

= H

⊕H

−

is a Pontryagin space with the

indeﬁnite scalar product (B

r, ρ)=(



Br,ρ).

Inversion of the Lagrange Theorem 701

Next, we introduce the space Y of the elements U =(u,ρ)

,withu ∈



J(F)

and ρ ∈ H -Ker



B satisfying the condition (3.3), i.e., by (2.7), ρ =Σ(u)+r, where

Σ(u)=

ωS



(x)·N (x)



u(z) ·η

(z)dz −P

(x) ·N (x)



u(z) ·η

(z)dz



r ∈ H

, P

= I −Q

.WealsosetJ



I 0





and J =ΠJ

Π where Π is the

orthogonal projection (in L

(F) × L

(G)) on the space Y .Let[U, V ] be a scalar

product in L

(F) ×L

(G). For arbitrary U ∈ Y we have

[JU, U]=[J

U, U ]=u

(F)

+(BΣu, Σu)+(BQρ,Qρ)

= u

(F)

+ S

−1

α=1



u · η

+(B

r, r).

The spaces Y

and Y

−

of the elements U

=(u, Σ(u)+r

)

and U

−

=(0,r

−

)

possess the properties [J(U

)] > 0, [J(U

−

)] < 0 for non-zero U

and

−

, respectively, moreover, they are J-orthogonal: [JU

−

] = 0. Finally, the

operator J is bounded, self-adjoint and invertible in Y :ifJU =0,then[JU, V ]=

0, ∀V ∈ Y ;takingV = U

−U

−

we obtain

0=[J

+ U

−

), (U

−U

−

)]

= u

(F)

+ S

−1

α=1



u ·η

+(B

) − (B

−

which implies U = 0. Hence Y is the Pontryagin space and [JU, V ] is an indeﬁnite

scalar product in Y .

Now we consider the expression [JA



U, U ]forU ∈ Y .

In view of Proposition 3, A



U ∈ Y, hence

[JA



U, U ]=[J



U, U ]=[A



U, J



(∇

u −∇(r

+ r

) − 2ω(e

× u)) · udx +



u ·N



B(¯ρ)dS

= −



|S(u)|

dx +2ω



¯u

− u

¯u

)dx +



(u · N



B ¯ρ −



Bρ

u ·N)dS.

Hence Re[JA



U, U ] ≤ 0, which means that −iA



is a J-dissipative operator

in Y . By the Pontryagin-M. Krein-Langer-Azizov theorem [5], A



has a ﬁnite-

dimensional invariant subspace L and the eigenvalues of A



have non-negative

real parts. We have seen above that the operator A



can not have imaginary eigen-

values (including λ = 0). Hence these eigenvalues have positive real parts.

If λ =0andU



=(u,ρ



)

satisﬁes the equation A



= λU



then U =(u,ρ)

with ρ = ρ



+ λ

−1

(u · N) satisﬁes (3.7) with the same λ. This completes the

proof of Theorem 1.

702 V.A. Solonnikov

4. The case of non-symmetric F

In this case the equation (1.2) deﬁnes one-parameter family of equilibrium ﬁgures

obtained by rotation of the angle θ of one of them, F

, around the x

-axis. In

what follows we mean by F arbitrary such ﬁgure. As shown in [13], a linearized

problem (1.10) takes the form

+2ω(e

×v) − ν∇

v + ∇p =0,

∇·v(x, t)=0,x∈F,t>0,

T (v,p)N + NB

ρ =0, (4.1)

= N(x) · v(y, t) −

h(x)



(z)dS



h(y)v(y, t) · N (y)dS, x ∈G,

v(x, 0) = v

(x),x∈F,ρ(x, 0) = ρ

(x),x∈G,

where

h(x)=η

(x) ·N(x).

In the case of axially symmetric F we have h(x) = 0, and the term with the

integral in the boundary condition drops out. The orthogonality conditions (1.11),

(1.12) should be supplemented with



ρ(x, t)h(x)dS =0.

By (2.5), Bh =0.

The operator A is deﬁned as above but the deﬁnition of its domain D(A)

contains the additional orthogonality condition



ρ(x)h(x)dS =0. (4.2)

Theorem 1 takes the following form:

Theorem 1



. Let

min

|θ|≤π



((x

cos θ + x

sin θ)

− x

)dx ≥ c

> 0. (4.3)

If the quadratic form (1.18) takes negative values for some ρ satisfying the condi-

tions (1.14), (4.2), then the operator A has a ﬁnite number of eigenvalues with a

positive real part.

Since h ∈ Ker



B, the proof of this theorem reduces to the analysis of the

spectrum of the operator A



, that is carried out as above. We prove the analogue

of Proposition 2.

Proposition 2



. Arbitrary ρ ∈ L

(G) can be represented in the form (2.7),whereρ

satisﬁes the conditions (2.2) and

(x)=a

(x) ·N(x)+a

(x) ·N(x).

Inversion of the Lagrange Theorem 703

The constants a

are found from the algebraic system

β=1



(y) · N(y)dSa



ρ(y)y

dS, α =1, 2. (4.4)

If ρ ∈ H,thenρ

∈ H

.Ifρ = η

(x) ·N(x), β =1, 2, then ρ

=0.

Proof. We turn to the equations (2.6):

(η

·N,x



−x

)dx ≡ B

(η

·N,x



−x

)dx ≡−A,

(η

·N,x

)=−(η

·N,x



dx ≡ C.

It follows that



(B(η

·N), η

· N)



α,β=1,2

= ω



B −C

−CA



. (4.5)

This matrix is positive deﬁnite, because, by (1.17),

Bξ

−2Cξ

+ Aξ





−x

)

−x

(ξ

+ ξ

)



dx ≥ c

(ξ

+ ξ

Hence its determinant is positive. The determinant of the matrix



, η

·N



α,β=1,2

= ω



C −A

B −C



is negative, and the system (4.4) for a

is uniquely solvable. The proposition is

proved.

As above, we have

(Bρ, ρ)=(Bρ

,ρ

)+(Bρ

,ρ

By the positivity of the matrix (4.5),

(Bρ

,ρ

α,β=1

(Bη

· N, η

· N)a

≥ c(a

+ a

) ≥ c(I

(ρ)+I

(ρ)).

It is also clear that

(Bρ

,ρ

) ≤ c(I

(ρ)+I

(ρ)).

Further arguments related to the analysis of the spectrum of A



are essentially

thesameasinSec.3.IfA



= λU



where λ =0andU



=(u,ρ



)

,thenAU = λU

with U =(u,ρ)

ρ = ρ



+ λ

−1



(u ·N ) −





u · NhdS



This proves Theorem 1



704 V.A. Solonnikov

References

[1] V.A. Solonnikov, Instability of rotating ﬂuid,J.Math.Sci.166 (2010), 91–105.

[2] V.A. Solonnikov, On the stability of uniformly rotating viscous incompressible self-

gravitating liquid, J. Math. Sci. 152, No. 5 (2008), 4343–4370.

[3] V.A. Solonnikov, On a linear problem connected with equilibrium ﬁgures of a uni-

formly rotating viscous liquid, J. Math. Sci. 132, No. 4 (2006), 502–521.

[4] V.A. Solonnikov, Letter to the editor, J. Math. Sci. 135 No. 6 (2006), 3522–3528.

[5] M.G. Krein, G.K. Langer, Defect subspaces and generalized resolvents of an Hermit-

ianoperatorinthespaceΠ

, Funct. Anal. Appl. 5 (1971), 136–146.

[6] A.G. Kostiuchenko, A.A. Shkalikov, M.Yu. Yurkin, On the stability of a top with a

cavity ﬁlled with a viscous ﬂuid, Funct. Anal. Appl. 32, No. 2 (1998), 110–113.

[7] M. Padula, V.A. Solonnikov, A simple proof of the linear instability of rotating liquid

drops, Ann. Univ. Ferrara 54 (2008), 107–122.

[8] M. Padula, Free work identity and nonlinear instability in ﬂuids with free boundaries,

Recent advances in elliptic and parabolic problems, Proc. of the International con-

ference, Hsinchu, 16–20.2.2004, Ch.-Ch. Chen, M. Shipot, Ch.-Sh. Chen (editors),

World Sci. Publ., (2005), 203–216.

[9] V.A. Solonnikov, On the problem of evolution of an isolated liquid mass,J.Math.

Sci 124 (2004), 5442–5460.

[10] S.G. Krein, On oscillations of a viscous ﬂuid in a container, Doklady AN SSSR, 159

(1964), 262–265.

[11] N.D. Kopachevsky, S.G. Krein, Operator approach to linear problems of hydrody-

namics,vol.2,Birkh¨auser, 2003.

[12] I.C. Gohberg, M.G. Krein, Introduction to the theory of linear nonselfadjoint opera-

tors, Providence, R.I.: AMS, 1969.

[13] V.A. Solonnikov, On the stability of nonsymmetric equilibrium ﬁgures of rotating

viscous incompressible liquid, Interfaces Free Bound. 6 (2004), 461–492.

[14] M.Sh. Birman, M.Z. Solomiak, Spectral theory of self-adjoint operators in a Hilbert

space, Kluwer Acad. Publ., 1987.

V.A. Solonnikov

Steklov Mathematical Institute

Fontanka 27

191023 S. Petersburg, Russia

e-mail: solonnik@pdmi.ras.ru

Progress in Nonlinear Diﬀerential Equations

and Their Applications, Vol. 80, 705–717

 2011 Springer Basel AG

Questions of Stability for

a Parabolic-hyperbolic System

Gerhard Str¨ohmer

Abstract. We consider a model describing a situation in which a population

follows the density gradient of a nutrient that is produced at a spatially inho-

mogeneous rate of production and subject to diﬀusion. We show the stability

of the equilibrium solution.

Mathematics Subject Classiﬁcation (2000). Primary: 35M13,

Secondary: 35Q92, 35B35.

Keywords. Stability, parabolic-hyperbolic systems, chemotaxis.

1. Introduction

In the paper [3] Chen, Friedman and Hu consider the partial diﬀerential equations

+div(u∇φ)=G (u, φ) , (1.1)

− Δφ = F (u, φ) ,

or, equivalently,

+ ∇u∇φ = G (u, φ) − uΔφ, (1.2)

− Δφ = F (u, φ) .

on a bounded domain Ω with the boundary condition

∂φ

∂n

=0 (1.3)

on ∂Ω. These equations provide a model for a system in which certain cells of

density u move through the region Ω in the direction of the gradient of the density

φ of the nutrient nourishing them. Their rate of reproduction also may depend

on the cell density and the density of the nutrient. Models of this kind are often

referred to as Keller-Segel models. For a review of this topic see, e.g., [6] and [7].

The author gratefully acknowledges support from the Max-Planck-Institute for Mathematics in

the Sciences in Leipzig, Germany, during the completion of this paper.

706 G. Str¨ohmer

In most cases they take the form of a parabolic system of equations, while in our

case the ﬁrst equation in (1.2) is, if we assume φ to be given for the moment, a

hyperbolic equation. This is also, e.g., true in [4], but in contrast to [3] the second

equation there is elliptic.

The characteristic curves for the hyperbolic equation are parallel to ∂Ωowing

to condition (1.3), thus no boundary condition for u is needed. Examples of F and

G discussed in [3] are

G (u, φ)=μk (u) φ − γu,

F (u, φ)=δ − k (u) φ

with positive numbers γ,δ,μ and a function k (u), in particular

k (u)=

1+u

Under certain assumptions Chen, Friedman and Hu prove the existence of a long-

term solution and the local stability of certain equilibrium states of the system.

These equilibria are such that u and φ equal constant numbers u

∗

and φ

∗

, respec-

tively, in Ω. Constant equilibrium solutions (u

∗

,φ

∗

) must fulﬁll the equations

G (u

∗

,φ

∗

)=0,F(u

∗

,φ

∗

)=0, (1.4)

thus we have two equations for two unknowns. In this paper we will only consider

the stability of equilibria, therefore the global assumptions about F and G made

in [3] are of no signiﬁcance to us. Deﬁning

∂G

∂u

∂G

∂φ

∂F

∂u

∂F

∂φ

∗

,φ

∗

= J,

Chen, Friedman and Hu assume the following two conditions for the stability of

the equilibria.

1. All eigenvalues of J have negative real part, or ad − cb > 0, a + d<0,

2. a + cu

∗

< 0.

This means that, assuming Condition 2, any constant solution of the system

of partial diﬀerential equations is stable with respect to spatially non-constant

perturbations if it is stable with respect to spatially constant perturbations. Now

it would be interesting to allow F and G, in particular the production rates for

the nutrient, to depend on the position x ∈ Ω. Equations (1.4) would then become

G (u

∗

,φ

∗

,x)=0,F (u

∗

,φ

∗

,x)=0. (1.5)

In most cases these equations only have solutions depending on x, and then they

are no longer the proper equations for an equilibrium, and one has to consider the

equations

∇u

∗

∇φ

∗

= G (u

∗

,φ

∗

,x) − u

∗

Δφ

∗

−Δφ

∗

= F (u

∗

,φ

∗

,x)

(1.6)

instead. Unless φ

∗

is constant, the cells would still move around in such an equilib-

rium state. The case of constant φ

∗

is of interest, though, as one would expect it

Stability for a Parabolic-hyperbolic System 707

to make sense to have equilibria without any movement with a population density

varying with the density of nutrient production. In case k (u)=u, i.e.,

G (u, φ, x)=μuφ −γu,F (u, φ, x)=δ (x) − uφ. (1.7)

there are such equilibria with a variable u

∗

and a constant φ

∗

,asF = G =0is

equivalent to

∗

= δ (x) ,

μu

∗

= γu

∗

and

∗

(x)=

δ (x) ,

while for other k there may not be such equilibria.

For k (u)=u the Jacobian matrix is

∂G

∂u

∂G

∂φ

∂F

∂u

∂F

∂φ

∗

,φ

∗

μφ

∗

− γμu

∗

−φ

∗

−u

∗

δ (x)

−

δ (x)

its trace is −

δ (x), its determinant is μδ (x)anda + cu

∗

= −δ (x). This means

that if δ (x) > 0 everywhere, Conditions 1 and 2 are always fulﬁlled pointwise.

For the remainder of the paper we assume that the functions F and G belong

to C

as functions of all their variables, and that ∂Ω belongs to C

as well, that

G (u

∗

(x) ,φ

∗

,x)=F (u

∗

(x) ,φ

∗

,x) = 0 for a constant φ

∗

and a strictly positive

function u

∗

∈ C





, and that Condition 2 is fulﬁlled pointwise. The formal

linearization of equations (1.2) at such an equilibrium is the system

= −∇u

∗

∇ϕ + G

∗

,φ

∗

,x) v + G

∗

,φ

∗

,x) ϕ − u

∗

Δϕ,

− Δϕ = F

∗

,φ

∗

,x) v + F

∗

,φ

∗

,x) ϕ

(1.8)

for perturbations ϕ of φ

∗

and v of u

∗

. Before we begin stating our result, we express

equations (1.2) in terms of the variables v = u − u

∗

and ϕ = φ −φ

∗

. We have

+ ∇v∇ϕ = −∇u

∗

∇ϕ +



G (v, ϕ, x) − (v + u

∗

)Δϕ,

−Δϕ =



F (v, ϕ,x)

(1.9)

with



F (v, ϕ,x)=F (v + u

∗

,ϕ+ φ

∗

,x)and



G (v,ϕ, x)=G (v + u

∗

,ϕ+ φ

∗

,x).

Let us also rewrite the system (1.8) in the form U

= LU with U =

vϕ

LU =

−∇u

∗

∇ϕ + G

∗

,φ

∗

,x) v + G

∗

,φ

∗

,x) ϕ −u

∗

Δϕ

Δϕ + F

∗

,φ

∗

,x) v + F

∗

,φ

∗

,x) ϕ

introducing the notations

D (L)=



U =

vϕ

| ϕ (x) ∈ C





,v(x) ∈ C





∂ϕ

∂n

