· Contents 1 Hilbert’s Boundary-Value Problem in Simply-Connected Domains. Im-mediate...

MSMSE-RSFSR

Novosibirsk State University

S.N. Antontsev

Boundary Problems

with Discontinuos Boundary Conditions and itsApplcations

A dissertation in competition for the scientific degreeof Candidate of Physical and Mathematical Sciences

(PhD in Physics and Mathematics)

Scientific advisor:Professor V. N. Monakhov

Novosibirsk — 1967

MSMSE-RSFSR

Universidade Estatal de Novosibirsk

S.N. Antontsev

Os problemes de fronteira

com as condicoes limites descontinuas e suasaplicacoes

Tese para defender de grau de candidato a doutor em fisicas-matematicas ciencias(PhD in Physics and Mathematics)

Cientifico coordenador:Professor, doutor de fisicas-matematicas ciencias

V. N. Monakhov

Novosibirsk — 1967

Contents

1 Hilbert’s Boundary-Value Problem in Simply-Connected Domains. Im-mediate Applications 71.1 The study of a class of operators . . . . . . . . . . . . . . . . . . . . . . . 71.2 Hilbert’s problem with discontinuous boundary conditions . . . . . . . . . 151.3 Free-boundary problems of filtration through inhomogeneous anisotropic soil 26

2 Boundary-Value Problems for Vector-Functions. Problems in Two -Connected Domains 352.1 Hilbert’s problems for vector-functions . . . . . . . . . . . . . . . . . . . . 352.2 Riemann’s problems for vector-functions . . . . . . . . . . . . . . . . . . . 382.3 Riemann’s problem with shift (Gazeman’s problem ) . . . . . . . . . . . . 392.4 Problems in two-connected domains . . . . . . . . . . . . . . . . . . . . . . 46

3 Free Boundary Problems of Gas Dynamics 493.1 Transformation of equations of gas dynamics . . . . . . . . . . . . . . . . . 493.2 Statement of problems of gas dynamics . . . . . . . . . . . . . . . . . . . . 553.3 A priori estimates of the solution . . . . . . . . . . . . . . . . . . . . . . . 583.4 The existence theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.5 The proof of existence of a solution for a Ryabushinskii-type problem . . . 663.6 The problem of overflow of a heavy incompressible fluid over a spillway . . 71

-1

Introduction

The dissertation is devoted to the study of boundary-value problems for quasilinear ellipticsystems of 2m first-order equations, (m ≥ 1), with two independent variables and toapplications of such systems. Besides, the coefficients in the boundary condition admitdiscontinuities of the first kind.

Problems of this sort are interesting from both the theoretical and practical points ofview, because they are nonlinear problems for partial differential equations, and becausemany free-boundary problems of continuum mechanics (of gas dynamics, hydrodynam-ics, theory of filtration and elasticity) can be reduced to problems with discontinuousboundary conditions.

This was because of relations with the gas dynamics that V.N. Monakhov [1] studiedthe so-called mixed boundary-value problem for quasilinear elliptic system (m = 1) andproved for the first time the unique solv ability of some boundary-value problems withdiscontinuous boundary conditions.

The dissertation consists of Introduction and three Chapters. In Chapter 1 we studyoperators of the following form

Pnf = −Π(z)

π

∫K

∫ (f(t)

Π(t)(t− z)+

z2n+1f(t)

Π(t)(1− tz)

)dKt, n ≥ 0,

Pnf = −Π(z)

π

∫K

∫ (f(t)

Π(t)(t− z)+

t2n−1f(t)

Π(t)(1− tz)

)dKt, −k = n < 0, (0.0.1)

where

Π(z) = (z − zk)αk , K : |z| ≤ 1, |zk| = 1, 0 ≤ αk ≤ 1.

It is proved that the operators Pn possess the following properties: Pn are completelycontinuous while the operators Un = ∂Pn

∂ξare linear and bounded in the space of functions

f ∈ Lp(K), p > 2, besides, ∂Pnf

∂ξ= f and ||

Un f ||L2 = ||f ||L2 (where

Un f is the singular

part of Unf ).

1

2 Introduction

We also consider the operators Λk, (k = 1, 2), in the two-connected domain G : ρ ≤|ξ| ≤ 1 which have the properties analogous to those of Pn, and, given any f , solve theboundary-value problems

<e (ak + ibk) Λk f = 0, (0.0.2)

where

a1 + ib1 =

1, |ξ| = 1, ξ = eiγ, γ ∈ [0, 2π];−i, ξ = ρeiγ, γ ∈ [0, 2π]

(0.0.3)

a2 + ib2 =

1, ξ = eiγ, γ ∈ [0, 2π];1, ξ = ρeiγ, γ ∈ [0, π];−i, ξ = ρeiγ, γ ∈ [0, 2π].

(0.0.4)

In Chapter 1 Hilbert’s boundary-value problem for elliptic quasilinear system (m = 1)of the first-order equations is studed. The problem can be formulated in the followingway. To find a solution of the system of equations

wz + µ1(z, w)wz + µ2(z, w)wz + F (z, w) = 0 (0.0.5)

which satisfies the boundary condition

<e(a+ ib)w(t) = 0, t ∈ Γ, (0.0.6)

where (a + ib) have discontinuities of finite genus at the boundary points tk, (k = 1, s).Problem (0.0.5), (0.0.6) was completely studed by V.S. Vinogradov [4],[5], [6] in the case(a(t) + i b(t)) ∈ Hα for all t ∈ Γ. The unique solvability of a special boundary-valueproblem (0.0.5), (0.0.6) with discontinuous boundary conditions was first proved by V.N.Monakhov [1],[3], for the case where F (z, w) ≡ F (z), µ2(z, w) ≡ 0 and

a+ i b =

1, γ ∈ [0, π];−i, γ ∈ [π, 2π]

With the use of the operators Pn, introduced in §1, problem (0.0.5), (0.0.6) is reduced tononlinear integral equations for which the unique solvability is proved.

Under certain additional assumptions on the character of discontinuities in the bound-ary condition, the main result of this section can be formulated as follows: problem (0.0.5),(0.0.6) always has a solution in the class of functions bounded at the points of disconti-nuity and such that within this function class κ = ind[a+ ib]Γ ≥ 0; the solution contains2κ + 1 arbitrary real constants. If κ < 0, then problem (0.0.5), (0.0.6) has a solution if2κ− 1 solvability conditions are fulfiled.

At the end of the section we give a condition sufficient to construct solutions un-bounded at separate points.

In §3 new existence and uniqueness theorems are proven for some free-boundary prob-lems of filtration in inhomogeneous anisotropic soil posed in unbounded domains. These

Introduction 3

problems admit direct application of results of §2. Analogous problems with boundeddomains of filtration were considered in work [2].

In Chapter 2 Hilbert’s boundary-value problem for the vector-valued functions whichstisfy a quasilinear elliptic system of 2m (m ≥ 1) equations. In the case where thecoefficients in the boundary condition are Holder-continuous and the system is linearsimilar problems were considered in works of B.V.Boyarskii [8, 9] and I.I.Danilyuk [11].We study the situation most frequent in applications when the coefficients of the matrix inthe boundary condition are peacewise constant and all the partial indexes of the problemκi are nonpositive.

If N =m∑k=1

(2|κi|−1) solvability conditions are fulfilled, and under certain assumptions

on the descontinuities in the boundary conditions, the uniqueness and existence theoremsare proved.

Riemann’s problem for quasilinear system of 2m (m ≥ 1) equations is considered in§2. This problem reduces to Hilbert’s problem considered in §1 but with a larger numberof unknown functions.

In §3 we consider Riemann’s problem with shift (Gazeman’s problem) for a completesystem (0.0.5) (m = 1): to find a bounded solution w = w(z) of the system of equations

L w = wz + µ1(z, w) wz + µ2(z, w) wz + F (z, w) = 0, (0.0.7)

posed on the whole of the plane (D+ +D− + Γ), which satisfies the following relations onthe boundary Γ:

w+[α(t)] = (a+ i b) w−(t). (0.0.8)

Here α = ϕ+i ψ ∈ C1α(Γ) is a complex-valued function such that |α′(t)| 6= 0 everywhere

on Γ, and, moreover, is a bijective mapping of Γ onto itself which preserves the index.

It is assumed that the coefficients µi, F, (a + ib) satisfy the conditions of §§1-2 ofChapter 1 ,throughout the domain of definition.

The solvability of the nonlinear problem (0.0.7), (0.0.8) is proved, the number ofsolutions is counted, and the solvability conditons are presented.

A more general conjugation problem was studied by L.G. Mikhailov [12] in the classof generalized analitic functions. To the best of our knowledge, Gazeman’s problem forthe nonlinear system (0.0.1) have not been considered yet even in the case of continuouscoefficients.

In the same section we study more complicated boundary-value problems with severalshifts in the smooth boundary conditions.

In §4, we make use of the operators Λ, introduced in Chapter 1, to prove the existenceand uniqueness theorems for the boundary-value problems (0.0.2) for system (0.0.5) intwo-connected domains.

4 Introduction

Analogous problems in the class of generalized analitic functions were studied byI.N.Vekua [13]. The free-boundary problems of hydrodynamics, studied in Chapter 3, arereduced to problems (0.0.1), (0.0.5).

Chapter 3 is devoted to the free-boundary problems of hydrodynamics and gas dy-namics. In §§1-4 we consider the following hydrodynamical problems posed in two-connected domains with free boundaries: the flow of an incompressible fluid in a chan-nel with a prescribed (or completely unknown) form of the walls around an unknown(or prescribed) obstacle. It is assumed that on the unknown (free) parts of the flowboundary (the channel walls or the unknown obstacle) the velocity distribution is given,q = q(x) ∈ C1

α, besides, q = q(x) 6= 0, q(±∞) = 1. On the prescribed parts of the bound-ary θ = θ(x) = arctg (dy/dx) is given, moreover, |θ| < π/2 − ε0, ε0 > 0, θ(±∞) = 0. Itis requested to define the form of the unknown parts of the flow boundary Dz, which isnot univalent, generally speaking, and the distribution of the univalent complex-valuedvelocity q(z) = e−iθ(z)

In §5 we study the problem of cavitational streamline of a system of non-simmetriccurvilinear arcs by a uniformly subsonic gas flow. The study is performed by Riabushin-ski’s scheme. Analogously to works [1, 3], by means of introduction of the new inde-pendent variable τ = x + iψ, (ψ is the stream function), the above-formulated problemsare reduced to boundary-value problems with disco ntinuous boundary conditions in aprescribed domain for the function w = w(τ) that satisfies the equation

wτ − µ(w) wτ = 0. (0.0.9)

The system, corresponding to equation (0.0.9), is elliptic in the sense of Petrovskii if thesought solution ω satisfies the inequalities

|<e w| < N <∞, |=m w| < π/2− ε0, ε0 > 0 (0.0.10)

everywhere in the flow domain.It is proven that for the solutions of the problems formulated in §2 a priori estimates

(0.0.10) hold throughout the flow domain. With the use of the intergal operators intro-duced in Section 1 of Chapter 1 the boundary-value problems for equation (0.0.9) arereduced to nonlinear integral equations. The solvability of these equations follows fromthe Lerey-Schauder fixed-point theorem for completely continuous operators.

For the problem of cavitational streamline, the uniqueness of the constructed solutionis a byproduct of results of §2 of Chapter 1.

In §6 we construct a numerical approximate solution to the problem of overflow of aheavy incompressible fluid over a spillway.

The results of the dissertation, in course of their receiving, were presented at The2-nd Conference of Mathematicians of Byelorussia in 1967, and also at the seminars ofthe Corresponding Member of the USSR Academy of Science L.V. Ovsyannykov in 1966,of the Corresponding Member of the USSR Academy of Science A.V. Bitsadze in 1967, of

Introduction 5

the Corresponding Member of the USSR Academy of Science N.N.Yanenko in 1967, andwere published in papers[14, 15, 16, 17, 18, 19].

In conclusion, I use the opportunity to express my sincere gratitude to the scientificdirector Valentin Nikolaevich Monakhov for his valuable advices and permanent attentionto this work.

6 Introduction

Chapter 1

Hilbert’s Boundary-Value Problemin Simply-Connected Domains.Immediate Applications

1.1 The study of a class of operators

1. Let us formulate some results of I.N.Vekua [13] and B.V. Boyarskii [10] on the prop-erties of the singular integral operators of the following class

Tf = T (ξ) = − 1

π

∫K

∫ f(t)

t− ξdKt; Sf =

∂Tf

∂ξ= − 1

π

∫K

∫ f(t)

(t− ξ)2dKt,

Tnf = T (ξ)− ξ2n+2T

(1

ξ

), (n ≥ 0),

Tnf = − 1

π

∫K

∫ [f(t)

t− ξ+t2k−1f(t)

1− tξ

]dKt, (−k = n < 0), (1.1.1)

Snf =∂Tnf

∂ξ= S f − 1

π

∫K

∫ (2n+ 1− tξ) ξ2n f(t)

(1− tξ)2dKt, (n ≥ 0),

Snf =∂Tnf

∂ξ= − 1

π

∫K

∫ [f(t)

(t− ξ)2+

t2kf(t)

(1− tξ)2

]dKt, (−k = n < 0), (1.1.2)

where K is the circle |t| ≤ 1.The operators Tnf ≡ Tn(ξ) possess the following properties:

• 1. ∂Tnf/∂ξ = f , if f ∈ Lp(K), p > 1;

7

8 Hilbert’s Problem

• 2. If f ∈ Lp(K), p > 2, the functions Tnf ≡ Tn(ξ) satisfy the Holder conditionwith the exponent γ = (p − 2)/p in the whole of the plane. Consequently, Tn arecompletely continuous in Lp, p > 2.

• 3. The operators Sn are linear and bounded in the function space Lp, p > 1 and

‖Sn ‖L2 = 1 (where

Sn is the singular part of Sn).

• 4. For each f ∈ Lp the function Tnf ≡ T (ξ) satisfies the boundary condition

<e[ξ−nTn(ξ)] = 0, ξ = eiγ, γ ∈ [0, 2π], n ≥ 0.

The function Tnf ≡ T (ξ), n < 0, satisfies the boundary condition

<e[ξ−nTn(ξ)] = 0, ξ = eiγ, γ ∈ [0, 2π], n < 0

only if the following equalities hold:

cm(f) = − 1

π

∫K

∫ [f(t)tm−1 + f(t)t2|n|−m−1

]dKt = 0, (m = 1, |n|).

2. Let us introduce the operators Pn

Pnf = Π(ξ)Tnf∗, f ∗ ≡ f(t)

Π(ξ),

where Π(ξ) =s∏

k=1(ξ− ξk)αk , and |ξk| = 1, 0 ≤ α < 1. In the case Π(ξ) = [i(ξ2− 1)]1/2 and

n = −1/2, the operator Pn was introduced and studied by V.N. Monakhov in [1]. Let usshow that the operators Pn possess the properties analogous to those of Tn. To this end,let us quote some results of Gegelia T.G [20] which will be needed later on.

Let Q, Q,Qk, (k = 1,m), be points of a bounded set of the n-dimensional Euclideanspace Rn, and r(Q, Q) be the distance between the points Q and Q.

Let us consider

ρ(Q) = Πm11 rαk(p−1)(Q, Q) · Πm

m1+1r−αk(Q, Q),

ρ∗(Q) = Πm11 rαk(Q, Q) · Πm

m1+1r−αk(Q, Q)

where 0 ≤ αk < n and Πsl =

s∏k=l

.

Let w(θ) be a bounded measurable function of points of the set E, besides, ∫σw(θ)dσ =

0, where σ is the surface of the unit sphere in Rn centered at the point Q. Accept thenotation Lρp = Lp(E, ρ(Q)), p > 1, for the space of functions f with the norm

‖f‖Lρp

=∫

E

∫|f |p · ρ · dE

1/p

.

Hilbert’s Problem 9

Then the operator

S∗ϕ(Q) =1

ρ∗(Q)

∫E

∫ w(θ)ρ∗(Q)

rn(Q, Q)w(Q) dE (1.1.3)

exists almost everywhere in E, and is a bounded linear operator in Lρp.The above-formulated result of Gegelia T.G implies the validity of the following lemma.

Lemma 1.1.1 The operator

SRϕ (Q) = |R(Q)|∫E

∫ w(θ)f(Q)

|R(Q)| rn(Q, Q)dE (1.1.4)

is a linear bounded operator in Lp(E), p > 1, where

R(Q) = Πm11 r−βk(Q, Qk) · Πm

m1+1r−βk(Q, Qk)

and

0 ≤ βk <

n/p, if k = 1,m1;nq

= n(p−1)p

, if k = m1 + 1,m

(Moreover, if the first of the products is absent, we set m1 = 0, and if the second one isabsent, then m1 = m).

The validity of the Lemma follows from the possibility to represent the operator SRf (Q)in the form (1.1.3) where in αk from entering the definitions of ρ and ρ∗ are calculated bythe formulas

αk =

pβk, if k = 1,m;p−1pβk, if k = m1 + 1,m

3. Denote Unf = ∂Pnf/∂ξ and, since f ∗(t) = f(t)/Π(t) ∈ Lp, p > 1, we have∂Tf ∗/∂ξ = Sf ∗ (see [13][Theorem 1.35]). Therefore

Unf =∂Pnf

π∂ξ=dΠ

dξTnF

∗ + ΠSnf∗, where f ∗ =

f

Π.

Calculating the derivatives, we now find out that

Unf =dΠ

dξ

∫K

∫ [f(t)

Π(t)(t− ξ)+

ξ2n+1f(t)

Π(t)(1− tξ)

]dKt

− Π(ξ)

π

∫K

∫ [f(t)

Π(t)(t− ξ)2+

xi)f(t)

Π(t)(1− tξ)2

]dKt,


Unf =dΠ

πdξ

∫K

∫ [f(t)

Π(t)(t− ξ)+

t2k−1f(t)

Π(t)(1− tξ)

]dKt

− Π(ξ)

π

∫K

∫ [f(t)

Π(t)(t− ξ)+

t2kf(t)

Π(t)(1− tξ)2

]dKt,

(−k = n < 0), where all the integrals are defined in the sense of the principal value.Let us show that Un are linear bounded operators in Lp(K), p > 2. Evidently, it is

sufficient to show for the operator

SΠf =

∂

∂ξ[Π(ξ)Tnf

∗], f ∗ =f

Π

As for the other summonds of the operator Un, the assertion will follow by construction(see [11]). The operator SΠ can be written in the following way:

SΠf = −ei argΠ

π

1−

s∑k=1

αk

SΠ

0 f0 −1

π

s∑k=1

αkei argΠkSΠ

k fk

where fk = f ei argΠk , Πk = Π(ξ)(ξ − ξk)−1, and the operators SΠ

k are represented in theform

SΠk = |Πk(ξ)|

∫K

∫ f(t)w(θ)

|Πk(t)||t− ξ|2dKt, (k = 0, s).

Here w(θ) = ei arg(t−ξ), besides,

2π∫0

w(θ)dθ =

2π∫0

e−2iθdθ = 0

along the circle |t − ξ| = 1. Thus, for ρ > 2 the operators SΠk satisfy the conditions of

Lemma 1.1.1 and, hence, are linear and bounded in Lp. The linearity and boundedness

in Lp of the operator SΠf =∂[ΠT ( f

Π]

∂ξis therefore proven because the presence of the factor

ei argΠk , whose modulus is bounded, does not affect the properties of SΠ.The operators Pn can be represented in the form

Pnf = −Π(ξ)

π

∫K

∫ f(t)

Π(t)(t− ξ)dKt + Pnf,

where Pnf are functions analitic in the circle |ξ| < 1 for all f ∈ Lp. By this reason, takinginto account that f ∗ = f/Π ∈ Lp′ , p′ > 1/m, and ∂

∂ξTnf

∗ = f ∗, we get:

∂Pnf

∂ξ= f ∈ Lp


On the other hand, as was proven above, Unf = ∂∂ξPnf ∈ Lp, p > 2. Hence, for each

f ∈ Lp(K), the operators Pn have the generalized derivatives with respect to ξ and ξ,which are summable in the circle K with the power p, p > 2. Hence, in accord with [21],

Pnf ≡ Pn(ξ) ∈ Cα(K), (0 < α < 1)

4. Let us calculate ‖Un‖L2 . We first consider Un (n ≥ 0). Adopt the following notation

A(ξ) = −Π(ξ)

π

∫K

∫ f(t)

Π(t)(t− ξ)dKt; B(ξ) = −Π(ξ)

π

∫K

∫ f(t)

Π(t)(1− tξ)dKt.

Then

Unf =∂A

∂ξ+ ξ2n+1∂B

∂ξ+ (2n+ 1)ξ2nB(ξ).

As was proven above, the operator (2n+ 1)2nB(ξ) is completely continuous, therefore wewill continue the norm of the singular operator

U0nf =

∂A

∂ξ+ ξ2n+1∂B

∂ξ

To begin with, let f ∈ C10(K), e.g., let f be a continuous differentiable function and f ≡ 0

for 1− ε ≤ |ξ| ≤ 1 (each f is corresponded by a proper ε). Then

‖U0nf‖2

L2=

(U0nf, U

0nf)

=

(∂A

∂ξ,∂A

∂ξ

)

+

(ξ2n∂B

∂ξ,∂A

∂ξ

)+

(ξ2n+1∂B

∂ξ, ξ2n+1∂B

∂ξ

)= E1 + E2 + E3 + E4. (1.1.5)

Apllying Green’s formula to the integrals over the domain and taking into account thatB(ξ) is analitic in the circle |ξ| < 1, we get:

E1 = (f, f) +1

2i

∫Γ

A∂A

∂ξdξ +

1

2i

∫Γ

Afdξ

= ‖f‖2L2

+1

2i

∫Γ

A∂A

∂ξdξ = ‖f‖2

L2+ J1, Γ = ξ : |ξ| = 1.

1

2i

∫Γ

Afdξ = 0 since f = 0 near the boundary

E2 = E3 =1

2i

∫Γ

Aξ2n+1∂B

∂ξdξ,


E4 =

(ξ2n+1∂B

∂ξ, ξ2n+1∂B

∂ξ

)≤(∂B

∂ξ,∂B

∂ξ

)=

1

2i

∫Γ

Aξ2n+1∂B

∂ξdξ

1

2i

∫Γ

B∂B

∂ξdξ = J4

The fact that equality (1.1.5) is real implies that Im (J1+E2+E3+J4) = Im (J1+J4) = 0.Thus, we only have to show that <e(J1 + E2 + E3 + J4) ≤ 0. Let us transform J1 + J4

making use of the following obvious relations between A and B

A = −ξBΠ

Π; B = −ξAΠ

Πon the boundary |ξ| = 1, (1.1.6)

and also of the analiticity of A and B (the latter is due to the fact that f is finitelysupported):

J1 + J4 =1

2i

∫Γ

[A∂A

∂ξ+ B

∂B

∂ξ

]dξ =

1

2i

∫Γ

|A|2∂ lnA

∂ξ+∂ lnB

∂ξ

dξ

=1

2i

∫Γ

|A|2 ddξ

lnAB dξ =1

2

2π∫0

|A|2d argABdγ

dγ

=1

2

2π∫0

|A|2 ddγ

(−ξΠ

Π

)dγ =

1

2

2π∫0

|A|2

2d argΠ

dγ− 1

dγ.

Let us calculate ddγargΠ.

d

dγln Π =

s∑k=1

αkξ − ξk

dξ

dγ=

d

dγln |Π|+ i

d

dγargΠ. (1.1.7)

Making equal the imaginary parts of (1.1.7), we get:

2id

dγargΠ =

s∑k=1

αk.

Thus, the inequalitys∑

k=1

αk ≤ 1 (1.1.8)

is a sufficient condition of nonpositivity of <e(J1 + J4).

Let us consider E2 fors∑

k=1αk = 1.

E2 =1

2i

∫Γ

Aξ−(2n+3)∂B

∂ξdξ =

1

π

∫ ∫K

f(t)

Π(t)

1

2πi

∫Γ

Π(ξ)ξ(2n+3)

ξ − t

∂B

∂ξdξ

dKt.


Continue ∂B∂ξ

outside the circle by the formula B∗(ξ) = ∂B(1/ξ)∂ξ

. The function B∗(ξ) isanalitic outside the circle and bounded at infinity, and we finally get that the functionB∗(ξ)Π(ξ)ξ−(2n+3) is univalent and analitic outside the circle and vanishes at infinity. Itfollows now from the Cauchy formula that E2 = E3 = 0. Q.E.D.

In the cases∑

k=1αk = 1− ε, ε > 0, a straitforward estimating of the integral in E2 (see

item 5, for instance), yields E2 = 0.

5. To calculate ‖Un‖L2 = ‖Un ‖L2 , (n < 0), let us take the integral over a circle of a

larger radius R (which is possible because for f ∈ C10 the functions Pn(ξ) ≡ Pnf and their

first derivatives are well-defined on the whole of the plane). Given an arbitrary R ≥ 1,we get then:∥∥∥U0

n

∥∥∥L2(K)

≥∥∥∥U0

n

∥∥∥L2(KR)

= ‖f‖2L2(KR) +

1

2π

∫Γ

Pnf∂Pnf

∂ξdξ = ‖f‖2

L2(KR) − JR,

where the function Pnf ≡ Pn(ξ), due to the finite nature of f , is defined on the whole of

the plane and is of the order |ξ|s∑

k=1

αk−1

at infinity.A direct estimating of the integral implies that

|JR| ≤ CR−ε → 0 when R→∞

where C is a constant independent of R. In the cases∑

k=1αk = 1, rewriting JR analogously

and using the Cauchy formula, we get: JR = 0.Thus, for f ∈ C1

0 , and under the assumption that inequality (1.1.8) is true,∥∥∥U0f∥∥∥L2

= ‖f‖L2 .

Hence, ‖U0‖L2 = 1 by virtue of the density of the family f ∈ C10 an L2 .

6. In the domain G : ρ ≤ |ξ| ≤ 1 of the variable ξ, let us consider the integraloperator T :

Tf = − 1

π

∫G

∫ f(t)

t− ξdGt.

According to [13], Tf ≡ T (ξ) ∈ C(p−2)/p(E) on the whole of the plane, and given any func-tion f ∈ Lp(G), p > 2. Moreover, the function Tf possesses the generalized derivatives∂ξTf = Sf, ∂ξTf = f , which are summable in G with the power p > 2.

Let us construct functions Φif , analitic in G and continuous in G , and solving the

following boundary-value problems:

<e Φ1f = −<e Tf = u(γ), |ξ| = 1, γ ∈ [0, 2π]

Im Φ1f = −Im Tf = v(γ), |ξ| = ρeiγ, γ ∈ [0, 2π],

(1.1.9)


<e Φ2f = −<e Tf = u(γ), |ξ| = eiγ, γ ∈ [0, 2π],

Im Φ2f = −Im Tf = v(γ), |ξ| = ρeiγ, γ ∈ [π, 2π],

<e Φ2f = −<e Tf = u0(γ), |ξ| = ρeiγ, γ ∈ [0, π],

(1.1.10)

Let us cut the ring by the positive part of the real axis, and then apply the conformalmapping ξ = ρe2πiz/ω1 , (ρ−1 = eπiω2/ω1), to map the obtained simple-connected domainonto the triangle with the sides ω1 and ω2/2 of the plane z.

With the use of the well-known formulas (see [22]), the functions Φif can be written

explicitly:

Φ1f [ξ(z)] =

1

πi

ω1+ω2/2∫ω2/2

u · g1(z, t)dt− i

ω1∫0

v · g1(z, t)dt

Φ2f [ξ(z)] =

1

πig(z)

ω1+ω2/2∫ω2/2

ug−1(t)g∗(z, t)dt+

ω1∫0

u0g−1g∗(z, t)dt

+ i

ω1∫ω1/2

vg−1g∗(z, t)dt

, g(z) =

√√√√σ(z − ω1/2)

σ(z),

g∗(z, t) = g1(z, t)− g0(0, t), g1(z, t) =σ(t− z + ω1/2)e−η(t−ξ)/2

σ(ω1/2)σ(t− z).

Here η = const, and σ = σ(z) is Weierstrass’s function with the periods ω1 and ω2.By virtue of the properties of Tf , u, v, v ∈ C(p−2)/p when f ∈ Lp(G), p > 2, therefore

Φif are Holder-continuous. Moreover, for the derivatives dΦi

f/dξ the following estimateshold: ∣∣∣∣∣dΦ

if

dξ

∣∣∣∣∣ ≤ Φ∗ε−2/p, (i = 1, 2), ξ∈O(±ρ), (1.1.11)

where |Φ∗| < M <∞, and ε is the distance from the boundary of the domain. Analogousestimates hold for the functions |dΦ2

f/dξ| in neighborhoods of the points ±ρ:∣∣∣∣∣dΦ2f

dξ

∣∣∣∣∣ ≤ Φ∗

|ξ ± ρ|1/2. (1.1.12)

Let us introduce the operators Λif = Tf + Φif . By construction, Λif ≡ Λi(ξ) ∈

C(p−2)/p, (2 < p < 4), and possess the generalized derivatives in ξ and ξ summablewith a power, besides, the derivatives satisfy the equality ∂Λf/∂ξ ≡ f and solve thehomogeneous problems (1.1.9) , (1.1.10).

It follows from the properties of the operator Tf and the estimates (1.1.11), (1.1.12)that the operators Ξif = ∂Λif/∂ξ are linear bounded operators in the space of functions


f ∈ Lp, p > 2. Let us calculate the L2-norms of the operators Ξi. To begin with, assumethat f ∈ C1

0 and that f = 0 near G−G. Proceeding by analogy with item 5 and applyingGreen’s formula in the two-connected domain G, we get:

‖Ξif‖2L2

= ‖f‖2L2

+1

2i

∫ΓG

sΛif∂Λif

∂ξdξ = ‖f‖2

L2+ Ji. (1.1.13)

Let us calculate Ji. Because of the finite nature of the function f , the functionsΛif ≡ Λ(ξ) are analitic in vicinity of the boundary and continuous together with theirfirst derivatives, therefore ∂Λif/∂ξ = dΛif/dξ, and Ji can be rewritten in the followingway:

Ji =1

2i

∫ΓG

ΛifdΛif

dξdξ =

1

2i

∫ΓG

ΛifdΛif

=1

2

∫Γg

(<eΛif) d (ImΛif)− (ImΛif) d (<eΛif) = 0.

This is because either <e Λif = 0 and, hence, d (<eΛif) = 0, or

Im Λif = d (Im Λif) = 0.

Thus, ‖Ξi‖L2= 1, since the function family is dense everywhere in L2.

1.2 Hilbert’s problem with discontinuous boundary

conditions

1. Let us consider a complete elliptic system of the first-order equations written as asingle complex-valued equation

wz + µ1(z, w)wz + µ2(z, w)wz + F (z, w) = 0, (1.2.1)

where w = w(z, z) = u+ iv, z = x+ iy,

∂

∂z=

1

2

(∂

∂x− i

∂

∂y

),

∂

∂z=

1

2

(∂

∂x+ i

∂

∂y

).

The coefficients µi(z, w), F (z, w) of the system are defined for all z ∈ K, (without lossof generality the domain K can be thought of as the unit circle), and any w from theextended complex domain.

With respect to the coefficients of the system, it is assumed the fulfillment of thefollowing conditions.


1.|µi(z, w1)− µi(z, w2)| ≤ N |w1 − w2|, (i = 1, 2)

2.|F (z, w1)− F (z, w2)| ≤ f(z)N |w1 − w2|, (i = 1, 2),

where N = const is independent of z, f ≥ 0, and ‖F (z, w)‖Lp, ‖f(z)‖Lp

≤M <∞for any fixed w and, moreover, there exists limits of µi and F when w →∞.

System (1.2.1) is assumed to be uniformly elliptic e.g.

|µ1(z, w)|+ |µ2(z, w)| ≤ µ0 < 1. (1.2.2)

Hilbert’s problem consists in finding a bounded solution to system (1.2.2) satisfyingthe following boundary condition:

Re(a+ ib)w(t) = 0. (1.2.3)

(Evidently, in the case F 6≡ 0 the homogeneuity of the boundary condition doesn’t reducethe generality of the problem).

It is assumed that a + ib ∈ Hα, (0 < α ≤ 1), for t ∈ [tk, tk+1], k = 1, s − 1, and hasdescontinuities of the first kind at the points tk, i.e., a(tk−0)+ib(tk−0) 6= a(tk+0)+ib(tk+0),(k = 1, s − 1), besides, a2(tk ± 0) + b2(tk ± 0) 6= 0. According to [23, 24], in the class ofanalitic functions the canonical function of the homogeneous problem (1.2.3) is given bythe formula

X(z) =1

2[Φ(z) + Φ ∗ (z)] , (1.2.4)

where Φ(z), Φ ∗ (z) are the canonical functions of the conjunction problems

(a+ ib)Φ+(t) + (a− ib)Φ−(t) = 0,(a+ ib)Φ∗+(t) + (a− ib)Φ− ∗ (t) = 0.

(1.2.5)

These functions are given by the formula (see [23])

Φ(z) = ceΓ(z) if |z| < 1,Φ(z) = cz−κeΓ(z) if |z| > 1,

(1.2.6)

where c is an arbitrary constant,

Γ(z) =1

2πi

∫Γ

ln[−t−κ a−ib

a+ib

]t− z

dt =1

2π

∫Γ

arg[−t−κ a−ib

a+ib

]t− z

dt (1.2.7)

(κ is the index of problem (1.2.5) is the prescribed class). Calculating Φκ(z), we now findout that


Φκ(z) =c

ceiαzκΦ(z), (1.2.8)

where

α =1

2π

∫ 2π

0arg

[−t−κa− ib

a+ ib

]dγ, t = eiγ.

Letting c = e−iα/3, we get from this that

X(z) = Φ(z)[zκ + 1]c (1.2.9)

or, for |z| = 1,

X(z) = cΦ(z)zκ/22<e[zκ/2 + z−κ/2

].

This means that the function

X(z) = cΦ(z)zκ/2 (1.2.10)

can be taken for the canonical function of problem (1.2.3) with an even κ. Representation(1.2.10) makes no sense for any odd κ.

Now consider the function

γ(z) =1

2iπ

∫Γ

θ

t− zdt; θ = arg

[−a− ib

a+ ib

](1.2.11)

It follows from the well-known properties of the Cauchy integrals, [23], that

γ(z) = βk ln(z − tk) + γ0(z) (1.2.12)

near the points tk, (k = 1, s), where γ0(z) is a holomorphic function bounded in neigh-borhoods of the points tk and tending to a limit as z → tk, while

βk =θ(tk − 0)− θ(tk + 0)

2π. (1.2.13)

According to [23, 24] the canonical function X which solves problem (1.2.3) admitsthe following representation in t he class of functions univalued in the circle |z| < 1 andbounded at the points of descontinuity tk, (k = 1, s):

X(z) =′∏(z − tk)e

γ(z) =s∏

k=1

(z − tk)αkR(z)zn, (1.2.14)

(the symbol ”prime” means that the product includes only the factors with tk correspond-ing to βk < 0), lnR(z) ∈ Hα, 0 < α ≤ 1; n is a positive or a negative number (the caseof n multiple to 1/2 will be considered specially).


In what follows, we consider problems (1.2.1), (1.2.2) whose canonical functions Xsatisfy condition (1.1.8) of Section 1. By (z − tk)

αk we mean the branch golomorphic inK, i.e., on the plane cut along Γ.

Let us perform the change of variables w = iR(z)w0(z). Since the function R(z) isanalitic and lnR(z) ∈ Hα, it has the generalized derivative summable with the powerp = 2

1−α > 2. For w0(z) we get then problem (1.2.1), (1.2.3) whose canonical function isthe function X(z) = i

∏sk=1(z − tk)

αkzn = −iΠ(z)zn. The properties of the coefficients ofsystem (1.2.1) remain the same.

By this reason, in what follows problem (1.2.1),(1.2.3) is considered under the assump-tion that the above transformations are already performed.

In dependence on the sign of n, the following statesments of problem (1.2.1), (1.2.3)are considered.

Problem I. (n ≥ 0) A solution w = w(z) to system (1.2.1) is sought which possessesthe generalized derivatives in the sense of Sobolev wz, wz summable with the power p > 2,w ∈ W 1

p and satisfies the boundary condition (1.2.3) with the additional relations

∫Γ

w(z)

Π(z)zkdθ, z = eiθ, (k = 0, 2n).

Problem II. (−k = n ≤ 0) We are seeking a solution of system (1.2.1) ω(z) ∈ W 1p

satisfying condition (1.2.3).

Let us formulate the auxiliary problem needed in what follows.

Problem II*. (n ≥ 0) It is requested to find a solution of system (1.2.2) w ∈ W 1p and

2k − 1 real constants which satisfy the condition

<e ((a+ ib)w(t)) = (a+ ib)Π(t)t−k<ek−1∑m=1

(lm + ih)tm + l0

.

2. We shall reduce problems (I), (II), (II*) to nonlinear singular integral equationsand prove their unique solvability. The solutions to problems (I), (II), (II*) will be soughtin the form w = Pnf (n ≥ 0 corresponds to problem I, n < 0 corresponds to problemsII, II* ), where Pn are the operators introduced in Section 1. It is easy to see that,by construction, Pnf , (n ≥ 0), satisfy the homogeneous boundary condition and theadditional relations for any function f ∈ Lp. Indeed, we have that on the boundary ofthe circle |z| = 1

w(z) = Pnf = −iΠ(z)zn

π2=

∫K

∫ (f(t)

Π(t)(t− z)zn

)dKt (1.2.15)


and since <e [(a+ ib)iΠ(z)zn] |Γ = 0, then <e [(a+ ib)w(z)]|Γ = 0, whence it follows thatw satisfies the boundary condition (1.2.3).

Let us multiply by 12πi(z−ξ) , z ∈ K, the both sides of the equality

w(z)

Π(z)= − 1

Π

∫K

∫ [f(t)

Π(t)(t− z)+

z2n+1f(t)

Π(t)(1− tz)

]dKt, (1.2.16)

and take the integral over the boundary of the both sides of (1.2.16).Taking into account the fact that the first term on the right-hand side of (1.2.16) is

analitic ouitside the circle while the second one is analitic inside, according to the Cauchytheorem and to ([13], p. 70), we get that for each f ∈ Lp(K) , p > 2,

1

2Π

∫Γ

w(z)dz

Π(z)(z − ξ)=ξ2n+1

π

∫K

∫ f(t)

Π(t)(1− tξ)dKt

Expanding the left and right sides of the last equality in the powers of ξ, |ξ| < 1, wearrive at the equality ∫ w(z)

Π(z)z−kdθ = 0, z = eiθ, (k = 0, 2n). (1.2.17)

The function w = w(z), satisfying relations (1.2.17), for |z| = 1 can be given therepresentation

w(z) = iΠ(z)zn2=m[zn+1Φ(z)

],

where Φ(z) is an analitic function with a nonnegative index and, hence, (zn+1Φ(z)) has notless than (2n+ 2) zeros at the boundary. It is not difficult to get the following expansionfor the function w = Pnf , (n < 0), (see [13], p.298):

zkPnf =Π(z)

π

(∫K

∫f ∗ tk−1dKt + z

∫K

∫ (f ∗ tk−2 + f ∗ tk

)dKt

+ . . .+ zk−1∫K

∫ (f ∗+f ∗ t2k−2

)dKt

]+ Π(z)T0(t

kf∗), f∗ =f

Π,(1.2.18)

where T0, Tn, (n ≥ 0) are the operators introduced in Section 1. Hence, <eT0f = 0 (seesubsec 1, Section 1).

Thus, Pnf , n ≥ 0, for every f ∈ Lp satisfy the boundary condition of Problem II∗:one only has to take for lm + ihm the coefficients from the expansion (1.2.18). The samefunction satisfies the boundary condition of Problem II only if the following 2k − 1 real-valued equalities are fulfilled:

− 1

π

∫K

∫ (f(t)

Π(t)tm−1+

f(t)

Π(t)t2k−m−1

)dKt = 0 (1.2.19)


It is easy to see that Pnf are univalent functions and satisfy the same conditions forn multiple to 1/2.

Let us consider the following mixed boundary-value problem of an arbitrary index,when a+ ib, is constant. To be precise, where

<e (znw(z)) = 0, z = eiγ, γ ∈ [0, π]=m (znw(z)) = 0, z = eiγ, γ ∈ [π, 2π].

(If n = 0 we get the usual boundary-value problem). According to the previous consider-ations, we have:

w(z) = −

√i(z2 − 1)

π

∫K

∫ f(t)√i(t2 − 1)(t− z)

+z2nf(t)√

i(t2 − 1)(1− tz)

dKt, n ≥ 0,

w(z) = −

√i(z2 − 1)

π

∫K

∫ f(t)√i(t2 − 1)(t− z)

+t2nf(t)√

i(t2 − 1)(1− tz)

dKt, n < 0

Substituting w = Pnf into equation (1.2.1), we obtain the following nonlinear singularintegral equations for the functions f ∈ Lp:

f + ∆nf = g, (1.2.20)

where

∆nf = µ1(z, Pnf)Unf + µ2(z, Pnf)Unf +DnPnf, g = −F (z, 0),

Dn =

F (z,Pnf)−F (z,0)

Pnf, if Pnf 6= 0;

0, if Pnf = 0.

Moreover, according to the properties of the functions F (z, w) (see §2, 1),

‖Dn‖Lp ≤M <∞.

3. Let us show that the norms of solutions to equation (1.2.20) are bounded by aconstant depending only on µ0, M , p. To this end, let us first prove the following lemma.

Lemma 1.2.1 The boundary-value problems (I),(II),(II*) for the linear equation

wz + µ1(z)wz + µ2(z)wz +D(z)w = 0, (1.2.21)

where |µ1|+ |µ2| ≤ µ0 < 1, ‖D‖Lp ≤ M <∞, admit only the trivial solution in the classof functions w(z) ∈ W 1

p (K), p > 2.


Proof. Let w(z) 6≡ 0 be a solution of the corresponding problem for equation (1.2.21),then w(z) is a solution of the following equation

wz + µ(z)wz +Dw = 0, (1.2.22)

where

µ = a1 + ib1 =

µ1(z) + µ2(z), if wz 6= 0,0, if wz = 0.

Let ξ = ξ(z) be a homeomorphism of the circle |z| < 1 onto the circle |ξ| < 1 satisfyingthe equation

ζz + µ(z)ζz +Dw = 0 (1.2.23)

and leaving the points 0, ±1 immovable.According to [10], such a homeomorphism always exists and possesses the generalized

derivatives which are integrable with the power p, where p > 2 depends only on µ0.Under this transformation, the equation for w = w[z(ξ)] takes on the form

wζ +D ∗ w = 0, (1.2.24)

where

D∗ =D[z(ξ)]

ξz(1− |µ|2).

Following [7], let us show that D∗ ∈ Lp′ , where 2 < p′ = p2

2(p−1)< p for p > 2. Let us

estimate ‖D∗‖Lp′.

‖D ∗ ‖Lp′=

∫ ∫Kζ

|D|p′dKζ

|ζz|p′ (1− |µ|2)p′

1/p′

≤ 1

1− µ20

∫ ∫Kζ

|D|p′dKζ

|ζz|p′1/p′

.

Passing in the last integral to the variable z and making use of the following expressionfor the Jacobian, Jζ = |ζz|2 − |ζz|2 = |ζz|2 (1− |µ|2), we get that

‖D∗‖Lp′≤ 1

1− µ20

∫ ∫Kζ

|D|p′|ζz|2 (1− |µ|2)|ζz|p′

dKζ

1/p′

≤ 1 + µ20

1− µ20

∫ ∫Kζ

|D|p′dKz

|ζz|p′−2

1/p′

≤ 1 + µ20

1− µ20

∫ ∫Kz

|D|p′rKz

1/p′r

dKz

|ζz|(p′−2)q

1/q(p′−2)

,

where r = 2(p−1)p

, q = 2(p−1)p−2

. (It is obvious that 1r

+ 1q

= 1 and p′r = p). Let us estimate

the second factor. Taking into account the formulas ζz = ζzJ−1z , Jz = |zζ |2(1 − |µ|2), we

obtain:


∫ ∫Kz

dKz

|ζz|(p′−2)q

=

∫ ∫Kζ

J1+(p′−2)qz

|ζz|p′−2qdKζ =

∫ ∫Kζ

(1− |µ|2

)1+(p−2)q|zζ |2+(p′−2)qdKζ

≤ (1 + µ20)

1+(p′−2)q∫ ∫

Kζ

|zζ |2+(p′−2)qdKζ = (1 + µ20)∫ ∫

Kζ|zζ |pdKζ .

Since zζ ∈ Lp, we have proved thereby that D∗ ∈ Lp′ , P ′ > 2. But then for every solutionof equation (1.2.11) the following representation is true [13]

ω(z) = Φ(ζ)eıT0[ıD∗], (1.2.25)

where

T0[ıD∗] = − 1

π

∫ ∫Kζ

ıD∗

t− z− rıD∗

1− tζ

dKζ ∈ C p′−2

p′, =miT0[iD

∗] = 0,

for |ζ| = 1, and Φ(ζ) = ω(ζ)e−ıT0[iD∗] is a function analitic in the circle |ζ| < 1 andsatisfying the Holder condition.

It follows from the representation (1.2.25) that indω(z)Γ ≥ 0. By virtue of (1.1.8),the right-hand side of the equality

<eeı argΠzαω

= |π(z)|<e

k−1∑m=1

(lm + ıhm) zm + l0

, |z| = 1,

must have not less than 2k zeroes on the boundary and, hence, lm = hm = 0. Therefore,ω = 0 in Problem II and, moreover, in Problem II*, since Φ(ζ) is a solution of thehomogeneous problem with a negative index, whence Φ(ζ) = 0.

If n ≥ 0, (the case of Problem I), every analitic function satisfying condition (1.2.3)(written in terms of the variable ζ), can be represented in the following way:

Φ(ζ) = Xζ(ζ)ζn

l0 +

n∑k=1

lk(ζk + ζ−2n−k

)+ ıhk

(ζk − ζ−2n−k

)= Xζ(ζ)P2n(ζ),

where Xζ(ζ)ıCk is the canonical function of the homogeneous problem.Since ω = ΦeıT0[iD∗] is a solution of Problem I, P2n has not less than (2n + 1) zeroes

on the boundary which means that P2n ≡ 0. The Lemma is thus proven.

4. Let us return to equation (1.2.20). By virtue of the assumptions made aboutthe coefficients of system (1.2.1), there exists a linear operator ∆′

n, which is called theasymptotic derivative, [26], defined by the equality

lim‖f‖Lp→0

‖∆′nf − δnf‖Lp

‖f‖Lp

= 0.


Namely, ∆′n = µ1Un + µ2Un + DnPn, where µi, D

n are equal to the limit values of thecoefficients µi(z, ω), D(z, ω) as ω →∞. Let us consider the linear equation

f + ∆′nf = (I + ∆′

n)f = g. (1.2.26)

Due to the continuity of ‖0

U ‖Lp with respect to p and to the condition of ellipticity ofequation (1.2.2), there can find out a p = 2 + δ(p), δ > 0, such that∥∥∥∥∥µ1

0

Un f + µ2

0

Un f

∥∥∥∥∥Lp

≤ µ0

∥∥∥∥ 0

Un

∥∥∥∥Lp

‖f‖Lp , (1.2.27)

where µ0‖0

Un ‖ < 1. This allows one to rewrite equation (1.2.26) in the equivalent form

f + Ynf = f +

(I + µ1

0

Un +µ2

0

Un

)−1

Anf =(I + µ1

0

Un +µ2

0

Un

)−1

g,

An =

−DPn − µ1[Un−0

Un]− µ2[Un−0

Un], n ≥ 0,

−DPn, n < 0,

where the operator Yn is completely continuous as a product of the completely continuous

operator An by the linear and bounded operator

(I + µ1

0

Un +µ2

0

Un

)−1

. Let f 6= 0 be

an arbitrary solution to equation (1.2.26). Then ω = Pnf is a solution to problem (I)(correspondingly, to problem II* with n < 0) for equation (1.2.21). On the other hand,it follows from Lemma 1.1.1 that if g ≡ 0, problems I,II and, correspondingly, equation(1.2.26) admit only the trivial solution. Hence, the nonhomogeneous equation (1.2.26) issolvable for every right-hand side, and there exists a constant ν,

ν = ν(µ0,M, p), such that∥∥∥(I + ∆′

n)−1∥∥∥Lp≤ ν.

Let us consider the nonlinear equation equivalent to (1.2.20)

(I + ∆′n)f = (∆′f −∆n)f + g (1.2.28)

and show that all its solutions are bounded in the norm of Lp, p > 2. Assume the contrary.Let ‖f‖Lp →∞, then, dividing (1.2.28), we get that

‖τ‖Lp =

∥∥∥∥∥ f

‖f‖Lp

∥∥∥∥∥Lp

≤ ν

‖∆′

nf −∆nf‖Lp

‖f‖Lp

+‖g‖Lp

‖f‖Lp

→ 0,

while ‖T‖Lp = 1. This proves that all solutions to equation (1.2.20) are bounded in thenorm of Lp, p > 2.


5. Let us show that equation (1.2.20) has at least one solution. Taking into accountinequality (1.2.27) one can rewrite equation (1.2.20) in the form

f = Ωn(ω) =

(I + µ1

0

Un +µ2

0

Un

)−1

Fn(z, ω),

Fn =

−F (z, ω), n < 0,

−F − µ1[Un−0

Un]f − µ2[Un−0

Un]f, n ≥ 0,

where

(I + µ1

0

Un +µ2

0

Un

)−1

is a bounded linear operator continuously depending on ω.

The operator Fn(z, ω) = Fn(z, Pnf) is completely continuous in Lp, p > 2 by virtue ofthe continuity of F (z, ω) with respect to ω, and due to the complete continuity of theoperators Pn. Let us show that the operator Ωn(ω) continuously depends on ω. Letω1 − ω2 → 0 and f = f 1 − f 2 = Ωn(ω1)− Ωn(ω2). Then f satisfies the equation

f + µ1(z, ω1)Unf + µ2(z, ω

1)Unf +F (z, ω1)− F (z, ω2)

ω1 − ω2Pnf = g,

where

g =[µ1(z, ω

2)− µ1(z, ω1)]Unf

2 +[µ(z, ω2)− µ2(z, ω

1)]Unf 2F (z, ω1)− F (z, ω2)

.

By the continuity of µi and F with respect to ω, we get that ‖g‖Lp → 0 as ω1−ω2 → 0,whence ‖f‖Lp → 0. Thus, the operator Ωn is completely continuous in Lp, p > 2, andmaps the whole of the space Lp(K) into a sphere of a finite radius r = r(µ0,M, p). Itfollows from the Schauder Theorem that the operator Ωn has at least one fixed pointf = f0 which corresponds to the solutions of problems (I), (II), (II*).

6. Let us show the uniqueness of solutions to problems (I), (II), (II*).Let ω = ω1 − ω2 be the difference of two solutions. Then ω satisfies the following

linear homogeneous equation of the type (1.2.21):

ωz + µ(z)ωz +B(z)ω = 0,

where

B =

µ1(z,ω2)−µ1(z,ω1)

ω2−ω1 ω2z + µ1(z,ω2)−µ1(z,ω1)

ω2−ω1 ω2z, ω2 6= ωω1,

0, ω2 = ω1;

µ(z) =

µ1 + µ2(ω2−ω1)z

(ω2−ω1)z

for ω1z 6= ω2

z ,

0 for ω1z = ω2

z .


It follows from the inequalities∣∣∣∣∣µi(z, ω2)− µi()z, ω1

ω2 − ω1

∣∣∣∣∣ ≤ N <∞, (i = 1, 2),

and ‖ω2z‖Lp < ∞ that B = B(z) ∈ Lp. To complete the proof one has to evoke the

uniqueness of solutions to problems (I), (II), (II*). Therefore, the followingtheorem isproven.

Theorem 1.2.1 Problems (I) and (II*) are uniquely solvable, and problem (II) is uniquelysolvable if the additional solvability conditions (1.2.19) are fulfilled.

If Problem (I) is considered without the additional conditions

∫Γ

ω(z)

Π(z)z−kdθ, z = rıθ, (k = 0, 2n),

then it is solvable, and its solution contains 2n+ 1 arbitrary constants.

5. Until now we were considering the narrowest class of solutions which were boundedon a closed domain. The problem of finding the solutions unbounded at separate points,either belonging to the interior of the domain or to its boundary, or being the pointsof discontinuity of the boundary condition, is also of significant interest. In the caseswhen µi(z, ω) vanishes at the pole, independently of ω, such solutions can be found bythe scheme that follows. For the sake of simplicity, let us consider the case n < 0 andF (z, ω) ≡ 0.

It is requested to find out a solution of the problem

wz + µ1(z, w)wz + µ2(z, w)wz = 0 (1.2.29)

having a pole of |n|-th order at the origin. To the previously made assumptions on thecoefficients of the problem we add the following one:

|µi(z, w)| ≤ µ0|z|α (1.2.30)

for z sufficiently small; α is an arbitrary positive number.We shall seek a solution to problem (1.2.29) in the form ω = znω0. For ω0 = ω0(z) we

get the following boundary-value problem:

ωz + µ1(z)ωz + µ2(z)ωz = 0 (1.2.31)

<e(a+ ıb)tnω0(t)

= h(t), t = eıγ, γ ∈ [0, 2π],


where

F (z, ω) = n

(µ1

z+ µ2

zn−1

zn· ωω

).

Moreover, taking into account (1.2.30), we get that ‖F (z, ω)‖Lp ≤ M < ∞. Thereby wearrive at the above-formulated boundary-value problem for the function ω0 = ω0(z).

Vanishing of µi in many applications is prescribed by the physical meaning of theproblem under consideration as it happens, for instance, in the problems of the theory offiltration through an inhomogeneous soil considered in §3.

1.3 Free-boundary problems of filtration through in-

homogeneous anisotropic soil

1. It is known, ([28]), that for the inhomogeneous anisotropic soil the piezometric pressureh(x, y) satisfies the following equation in the filtration domain

(∇ (K∇h)) = 0 (1.3.1)

where K = (Kij) is the symmetric filtraton tensor, i.e Kij = Kji, i, j = 1, 2 and

δ = K11K22 −K12 +K21

2

4= K11K22 −K12

2 ≥ δ0 > 0 (1.3.2)

and |Kij(x, y)| ≤M <∞.Introducing the stream function ψ = ψ(x, y) and the generalized potencial φ = −k0h,

(k0 = const), we arrive at the system of equations

−ψy +K011φx +K0

12φy = 0,ψx +K0

21φx +K022φy = 0,

(1.3.3)

where K0 =(Kij

k0

)is the reduced filtration tensor. The last system can be rewritten as

the single equation for the complex-valued function w(z) = φ+ iψ, z = x+ iy,

wz − µ1(z, w)wz − µ2(z, w)wz = 0, (1.3.4)

where

µ1 =K0

11 −K022 + 2iK0

12

1 +K011 +K0

22 + δ; µ2 =

1−K011K

022 +K0

12K021

1 +K011 +K0

22 + δ.

Moreover, it follows from conditions (1.1.2) that

|µ1|+ |µ2| ≤ µ0(δ0,M) < 1. (1.3.5)

Passing in (1.3.4) to the inverse function s, we get the equation for the functionz = z(w)


zw − µ1(z, w)zw − µ2(z, w)wz = 0. (1.3.6)

As for the boundary conditions, in filtration problems these conditions are the same that

in the problems formulated for homogeneous isothropic soil. Let the gravity be directedin parallel to the axis OY , and let the prescribed water-insulated parts of the boundary ofthe filtration domain Dz (where ψ = const), and the permeable walls (where φ = const),be rectilinear. On the unknown free boundary, which is the stream line ψ = const,either the habitual condition of the pressure constancy is posed, φ+ k0y = const, or theprofile of the filtration pressure is precribed, h = h(x) or h = h(y). Therefore, eitherφ = φ(x) or φ = φ(y) is given. Let the filtration domain Dz be unbounded and assumethat that the soil is homogeneous and isothropic in some bounded domain DR

z ⊂ Dz, i.e.,k = K(x, y, h) = 1 for |z| ≥ R , whence

|µi(z, w)| = 0, |z| ≥ R = 0 (1.3.7)

for all w and some fixed R < ∞. In filtration problems, the boundary conditions in

the domain of complex potential w = φ + ψ define the domain Dw which is the imageof the filtration domain Dz bounded by the straight lines φ = const and ψ = const.The equations of the prescribed straight portions of the boundary y + aix = bi and theconditions x = x(φ) or y = y(φ) posed on the free boundary allow one to formulate forthe function z = z(w) satisfying equation (1.3.6) Gilbert’s problem with discontinuousboundary conditions.

In the this work we bound ourselves by the study of the unique solvability of thefollowing filtration problems.

2. The problem of filtration in the ground dam upon a drained base (for the case ofinhomogeneous soil see [29], p.267).


The straight line y = 0 is the ground level where ψ = 0, at the bottom of the flow,e.g., at the half-lines y = y01 − (A′C) and y = y1 − (BC) we have φ = φ1 and φ = φ2,on the wall AH ′ we have y − tanαx = y0, and on the free boundary AB ψ = ψ0 andφ+k0(y−H) = φ1, besides, φ2 = φ1 +k0(H− y1). The domain Dw in this problem is therectangle Dw : φ1 ≤ φ ≤ φ2, 0 ≤ ψ ≤ ψ1. Let us fix, in an arbitrary way, the imageof the point A′(φA′ , ψA′) on the segment C ′A′ of the boundary of Dw, and then performa conformal mapping of the domain Dw onto the circle |ζ| < 1 by means of the functionw = G(ζ, φ, φ0, ψ1, H) so that the points C ′, C, and A′ correspond to the points −1, 1,ı. Let tAe

ıγA and tB = eıγB be the images of the points A and B on the circumference|ζ| = 1. For the function z = z(ζ) = z[w(ζ)] we get then the following boundary-valueproblem:

y − tanα = a(γ), γ ∈[γA,

π2

],

y = a(γ), γ ∈[γA,

π2

],

(1.3.8)

where a(γ) equals, correspondingly, 0 for γ ∈ [π, 2π], (CC ′), y0 for γ ∈ [γA, π], C ′A′A, y1

for γ ∈ [0, γB], (BC), and, finally, a(γ) = φ(γ)−φ0

k0− H for γ ∈ [γB, γA], (AB). Equation

(1.3.6) transforms then to the form

zζ + µ1zζ + µ2zζ = 0, (1.3.9)

where

µ2 = µ2, and µ1 = exp

−2 arg

dw

dζ

µ1.

Let us seek a solution to problem (1.3.6), (1.3.8) in the form

z = z(ζ) = ıPnf + Φ(ζ), n = − 1

2, (1.3.10)

where ıPnf satisfies the homogeneous condition (1.3.8)

Pnf =Π(ζ)

π

∫ ∫K

[f(t)

Π(t)(t− z)+

f(t)

Π(t)(1− tζ)

]dKt

for every function f ∈ Lp, and

Π(ζ) = (ζ − 1)α(ζ − 1)1−α.

The analitic function Φ(ζ) satisfies the boundary conditions (1.3.8) and can be explic-itly given by the following formula:

Φ(ζ) =Π(ζ)

πı

∫|t|=1

a0(t)

Π(t)

t+ ζ

t− ζdt, (1.3.11)


where

a0(t) =

a(γ), γ 6∈[γA,

π2

], t = eıγ,

− a(γ)ı+tanα

, γ ∈[γA,

π2

].

Obviously, (see, for instance, formulas (1.2.12) of §2), the function Φ(ζ) has logarythmicsingulariries at the points ±1 of discontinuity of a0(t).

Φ(ζ) = − y1

2πıln(ζ − 1) +

y0

2πıln(ζ + 1) + Φ∗(ζ),

where ∥∥∥∥∥dΦ∗

dζ

∥∥∥∥∥Lp

<∞, p > 2.

Substituting z = ıPnf + Φ into equation (1.3.6), we get the integral equation for definingthe function f

f + µ1Unf + µ2Unf − ıµ1dΦ

dζ− ıµ2

dΦ

dζ. (1.3.12)

Besides, by virtue of condition (1.3.7) on the functions µi∥∥∥∥∥µ1dΦ

dζ+ µ2

dΦ

dζ

∥∥∥∥∥Lp

≤ N

R, ∥∥∥∥∥dΦ∗

dζ

∥∥∥∥∥Lp

,where the constant N is independent of f , whence, analogously to §1, 4,

‖f‖Lp ≤N

1− µ0Λnp

;(Λnp = ‖

0

Un ‖ = ‖Un‖Lp = 1 + δ(p)).

According to §1, 5, equation (1.3.12) and, hence, problem (1.3.8), (1.3.9), both have atleast one solution

z = z(ζ, y0, y, ψ0, ψ1, φ1, H), w = G[ζ, ψ0, ψ, φ1, H].

Moreover, due to the regularity of the boundary functions we get that zζ , zζ ∈ Cα every-where exept neighborhoods of the points ±1, eıγA , eıγB . It follows now from Lemma 1.1.1and the solvability of the homogeneous problem (1.3.8) in the class of analitic functionsthat the solution to problem (1.3.8), (1.3.9) is unique.

To make the solution obtained a physically reasonable solution to the original filtrationproblem, it is necessary to claim that the Jacobian of the mapping J = D(x,y)

D(ζ,z)= |zζ |2−|zζ |2

is nonzero everywhere in the closed circle |ζ| ≤ 1 except, maybe, the special points ±1,eıγB , and eıγ on the circumference |ζ| = 1. Considering an auxiliary problem for thesought functions


θ = arctan∂y(φ, ψ)/∂φ

∂x(φ, ψ)/∂φand v =

∂y(φ, ψ)

∂φ

one can separate from the corresponiding family of solutions depending on the paranetersy0, y, φ1, ψ0, ψ,H those solutions which are admissible from the physical points of view.

3. The problem of filtration of a fluid from a pool ( drainage-basin) into a horizo ntallayer bounded by two impermeable layers. (Fig. 3).

On the stream lines ABC and A′C we have ψ = ψ0, and ψ = 0 on AB. Besides,φ+k0(y+H) = φ1, φ = φ1 on A′A and, moreover, φ2 = φ1 +k0(H−y). The problem hastherefore no principle differences with problem 2. and the proof of its unique solvabilityis performed in completely the same way.

4. The problem of defining the form of a hydrotechnical construction by the prescribedfrofile of the filtration pressure and the finite depth of the water-permeable layer.

The filtration problems of this type were considered for the first time in the case ofhomogeneous soil by P.Ya.Polubarinova-Kochina (see, for instance, [30, 29], p.224) , andlater on in works of M.T.Nuzhin and N.B.Il’inskii (see, for example, [31] ).

On the stream lines C ′C and AA′B we have ψ = 0 and ψ = ψ0 correspondingly, onthe free boundary one of the two conditions for φ = −k0h(x, y) is posed:

φ = f(x), x ∈ [0, xB], (1.3.13)


φ =

f1(y), y ∈ [y0, H], on A′A,f2(y), y ∈ [y0, y1], on A′B.

(1.3.14)

On the permeable portions C ′A and BC we have φ = φ1 and φ = φ2 respectively.Besides, the boundary conditions are assumed compatible for the condition (1.3.13), φ1 =f(0), and for the condition (1.3.14), f1(y0) = f2(y0). The functions fi, i = 1, 2, areassumed to be continuous differentiable with respect to their arguments

0 < m ≤∣∣∣∣∣dfidt

∣∣∣∣∣ ≤M <∞.

By means of a conformal mapping of the circle |ζ| < 1 onto the domain Dw, in whichthe points A, B, C correspond to the points ±1 and −ı, the boundary-value problems(1.3.13) and (1.3.14) are reduced to the problems

x = a(γ), γ ∈ [0, π],y = a(γ), γ ∈ [π, 2π],

(1.3.15)

y = a(γ), γ ∈ [0, 2π] (1.3.16)

for the differential equation (1.3.9). In the boundary condition (1.3.15) a(γ) = f−1[φ(γ)]for γ ∈ [0, π], and for γ ∈ [π, 2π] a(γ) is piecewise-continuous with the points of disconti-nuity γC and −ı. In condition (1.3.16), a(γ) = f−1

i [φ(γ)], (i = 1, 2), for γ ∈ [0, π], and ispiecewise-constant for γ ∈ [π, 2π] with the points of discontinuity γC and −ı.

The boundary-value problem (1.3.15), (1.3.9) can be reduced to an integro-differentialequation of the type (1.3.12) by means of the substitution z = eıπ/4Pnf + Φ(z), wherePn is defined by formula (1.3.10) with Π(ζ) ≡

√ζ2 − 1 and the analitic function Φ(ζ)

satisfies the boundary conditions (1.3.15).The rest of the proof of the unique solvability of problems (1.3.12), (1.3.9) and (1.3.15),

(1.3.10) is completely analogous to the proof given for problem (1.3.11), (1.3.9).

5. Let us consider now the case where a part of the filtration domain, say, AA′, iscurvilinear e.g. is given by the equaqtion f(x, y) = 0 (in 2, 3). The boundary condition(1.3.8) and its counterpart in problem 3 are substituted by the following nonlinear one:

f(x, y) = 0, γ ∈[γA,

π2

]y = a(γ), γ 6∈

[γA,

π2

].

(1.3.17)

Let us introduce the function Z = f(x, y)+ ıy. It is easy to calculate the Jacobia n of thetransformation:

J = |Zz|2 − |Zz|2 =|f ′x − ıf ′y + 1|2 − |f ′x + ıf ′y − 1|2

1

4=f ′x2. (1.3.18)


Hence, if f ′x > 0, then our transformation is bijective. Let us deduce the equation satisfiedby the function Z = Z(ζ) = Z[z(ζ)]. Differentiating the equality in ζ and ζ, we arrive atthe following relations:

Zζ = Zzzζ + Zzzζ ,Zζ = Zzzζ + Zzzζ .

(1.3.19)

Having found zζ and zζ from (1.3.19) and substituting them into (1.3.9), we obtain thefollowing equation for Z = Z(ζ):

Zζ + q1(ζ, Z)Zζ + q2(ζ, Z)Zζ = 0, (1.3.20)

where

q1 =(Zz)

2 − (Zz)2 − 2ıZzZz=mµ2

|Zz − µ2Zz|2 − |µ1Zz|2,

q2 =µ2 (|Zz|2 + |Zz|2 − µ2ZzZz) + ZzZz

(|µ1| − ZzZ

)|Zz − µ2Zz|2 − |µ1Zz|2

.

Moreover, |q1|+ |q2| ≤ q0 (|f ′x| , µ0) < 1 if 0 < |f ′x| <∞.

The boundary condition (1.3.17) transforms to the form

<eZ = 0, γ ∈[γA,

π2

],

=mZ = a(γ), γ ∈[γA,

π2

].

(1.3.21)

Therefore, we have arraived once again at problem (1.3.8) for equation (1.3.20) and,hence, if the curvilinear boundary is such that 0 < |f ′x| < ∞, problem (1.3.17), (1.3.9)has a unique solution.

6. The following problems of filtration in an inhomogeneous anisothropic soil with afinite domain of filtration can be studied completely analogous to the preceding consid-erations.

The problem of filtrating of a fluid from a pool with the free boundary arriving at thedrain.


The problem of filtration of a fluid from an axisymmetric well.

Restricting the study of the latter problem to the half of the filtration domain AA′BC,and performing the change of variables (1.3.18), in both cases we arrive at the followingproblem: it is requested to find out a solution of the equation

Zζ + q1Zζ + q2Zζ = 0, (1.3.22)

bounded in the circle |ζ| ≤ 1 and satisfying the boundary conditions (1.3.21), besides

a = a(γ) ∈ Cα.

According to results of §2, 5, problem (1.3.21), (1.3.22) has a unique solution.

Chapter 2

Boundary-Value Problems forVector-Functions. Problems in Two-Connected Domains

2.1 Hilbert’s problems for vector-functions

1. The problem is studied of finding a bounded vector-solution ω(z) = ω1, ω2, . . . , ωn ∈W 1p , p > 2, to the system of equations

ωz + µ1(z, ω)ωz + µ2(z, ω)ωz = g(z) (2.1.1)

by the boundary condition

<e [Ω(t)ω(t)] = 0 , t = eıγ, γ ∈ [0, 2π], (2.1.2)

where µi, Ω = ‖Ωij‖, are square matrixes of dimension n, g(t) = g1(t), g2(t), . . . , gn(t)is a vector, and ωi, Ωij, gi are complex-valued functions. (Without loss of generality weassume that z varies within the unit circle). We make the following assumptions aboutthe coefficients of system (2.1.1) and the matrix on the boundary condition. For everyz ∈ K µi(z, ω) satisfy the Lipschitz condition with respect to ω, i.e., all the elementsof the matrixes µi = ‖µkji ‖ satisfy this condition in ω. The matrix Ω = ‖Ωij‖ satisfiesthe Holder condition on Γ except the points tk, (k = 1, p) where its elements may havediscontinuities of the first kind, that is, Ωij(tk − 0) 6= Ωij(tk + 0). Moreover, detΩ nevervanishes on Γ including the points of discontinuity tk, (k = 1, p).

System (2.1.1) is assumed to be uniformly elliptic, i.e.,

maxi

m∑j=1

|µij1 |+ maxi

m∑j=1

|µij2 | ≤ µ0 < 1. (2.1.3)

Let us call the summary index of problem (2.1.1), (2.1.1) the quantity

35

36 Problems for vector-functions

κ = ind detΩ =m∑i=1

κi,

where κi are the partial indexes of the canonical system of solutions in the class of analiticfunctions bounded at the points of discontinuity (see [25, 32]). According to [23, 32], thecanonical matrix of solution to the homogeneous problem in the mentioned class can berepresented in the following way:

X(z) = ıX0(z)Zκ(z) = ıX0(z)

∥∥∥∥∥∥∥∥∥∥∥∥∥∥

zκ1

zκ2

···zκm

∥∥∥∥∥∥∥∥∥∥∥∥∥∥,

where

X0(z) =∥∥∥Πk=1n(z − zk)

βik

∥∥∥ = ‖Xij‖,

if the coefficients of the matrix ω are piecewise constant (which we still assume). detX 6= 0everywhere on γ except, maybe, the points of discontinuity. The solution of problem(2.1.1) -(2.1.2) will be sought in the form w = Tκf .

The operator Tκ maps every vector f = f1, . . . , fm into a vector of the same dimen-sion and has the form

Tκf = − X0(z)

π

∫ ∫K

[X0(t)−1f(t)

t− z+G(z, t)

X0(t)−1f(t)

1− tz

]dKt,

where

G(z, t) = ‖Gi(z, t)‖, Gi(z, t) =

z2κi+1, κi ≥ 0,

t2|κi|−1, κi < 0.

Let all κi ≥ 0, then for every function f = f1, . . . , fm ∈ Lp the vector ω(z) = Tκfsatisfies condition (2.1.2). Indeed, taking into account that G(z, t) = ‖z2κi+1‖ = Z(z),and that Zκ(z) = Z−κ(z) on the circumference |z| = 1, we get that

Tκf = − ıX0(z)Zκ

π

∫ ∫K

2=m[Z−κX0(t)−1 f

t− z

]dKt for |z| = 1.

If some κi < 0, (let these are κi with i = p,m), Tκf satisfies the homogeneous condition(2.1.2) only if the following N =

∑mi=p(2|κi| − 1) conditions are fulfilled:

Cji (f) = − 1

π

∫ ∫K

[f ∗i t

j−i + f ∗i t2|κi|−j−1

]dKt = 0, (2.1.4)

Problems for vector-functions 37

(j = 1, |κ|), i = p,m,

where f ∗i are the components of the vector f ∗ = [X0(t)]−1f(t). It follows directly from the

structure of the operator Tκf that ∂Tf∂z

= f . Applying the results of Chapter 1, Section1 to each of the compo nents of the vector Sκf = ∂

∂zTf we conclude that S is a linear

operator and is bounded in the function space f ∈ Lp(K), p > 2.

Let us now estimate ‖Sκf‖2L2

=(Sκf, Sκf

)=∑mk=1 ‖f ∗‖L2 , where f ∗ are the compo-

nents of the vector Sκf .

Let us consider first the case where all κi ≤ 0 which occurs the most frequentlyin problems with discontinuous boundary conditions. Then, analogous to Section 1 ofChapter 1, we have for f ∈ C1

0(K)

‖Sκf‖2L2≤ ‖f‖2

L2+

1

2ı

∫ΓR

Tκf∂Tκf

∂ζdζ = ‖f‖2

L2+ JR, (2.1.5)

where the expression Tκf∂Tκf∂ζ

is understood as the scalar product of the correspondingvectors.

Treating JR and making use, as in Section 1 of Chapter 1, either of the Cauchytheorem, or of a direct estimate of the integral, we get that under the condition (1.8) ofSec.1, Ch.1 JR → 0 when R→∞. Thus, ‖Sκ L2| = 1 if all κi ≤ 0.

3. Substituting into equation (2.1.1) ω = Tκf we get the singular integral equationfor the vector f

f + µ(z, ω)Sκf + µ2(z, ω)Sκf = g(z). (2.1.6)

Due to the continuity of Λκp = ‖Sκ‖Lp with respect to p, and by virtue of condition (1.1.8),

there exists p > 2 such that

∥∥∥µ1Sκf + µ2Sκf∥∥∥Lp≤ µ0Λ

∗p‖f‖Lp ,

where µ0Λκp < 1 for p close enough to 2 and, hence, to equation (2.1.6) the principle of

contraction mappings is applicable.

Thus, there holds

Theorem 2.1.1 Equation (2.1.6) has a unique solution f , which corresponds to theunique solution of problem (2.1.1), (2.1.2) κi ≤ 0, only if the solvability conditions (2.1.4)are fulfilled.


2.2 Riemann’s problems for vector-functions

The Riemann problem for a quasilinear elliptic system of 2m, (m ≥ 1), can be formulatedas follows: it is requested to find a vector-solution ω = ω1, . . . , ωm to the system ofequations, posed on the whole of the plane,

Lω = ωz + µ1(z, ω)ωz + µ2(z, ω)ωz = 0, (2.2.1)

whose values on the boundary Γ satisfy the relation

ω+(t) = (a+ ıb)ω−(t) + g(t), t = eıγ ∈ Γ, (2.2.2)

where µi(z, ω), a+ ıb are square matrixes of dimension m, ω± ∈ W 1p , p > 2, are the values

of the vector ω in the domains D± respectively.

As usual, we assume that the domain D+ is the circle |z| < 1. For µi, a + ıb theassumptions of Section 1 are supposed to be true.

Let us take into consideration the m-component vector-valued function ω∗ = ω∗(z)deifined for all z ∈ D+ as follows:

ω∗(z) = ω−(

1

z

). (2.2.3)

It is obvious that z = 1z, ω∗(t) = ω−(t) on the boundary. For the 2m-component vector

W = ω+(z), ω∗(z), defined for z ∈ D+, we then arrive at the following Hilbert’s problemalready considered in Section 1.

Wz + q1(z,W )Wz + q2(z,W )W z = 0,<eΩ(t)W (t) = h(t), t = eıγ, γ ∈ [0, 2π],

(2.2.4)

where the block-diagonal matrixes qi , Ω of dimension 2m× 2m have the form

qi(z,W ) =

∥∥∥∥∥ µ+i (z, ω+) 0

0 z2

z2µ−i

(1z, ω∗

) ∥∥∥∥∥ , Ω =

∥∥∥∥∥ I a− ıbI ı(a− ıb)

∥∥∥∥∥and h(t)<e g,=mg.

Given a solution to problem (2.2.4), the solution to problem (2.2.1), (2.2.2) is definedby the vector-function

ω(z) =

ω+(z), z ∈ D+,

ω∗(

1z

), z ∈ D−

We proceed to consider in more detail Riemann’s problem with shift for the completesystem of equations with m = 1.


2.3 Riemann’s problem with shift (Gazeman’s prob-

lem )

1. Riemann’s problem with shift (Gazeman’s problem) for the quasilinear elliptic systemof equations (2.2.1) can be formulated in the following way: it is requested to find abounded solution to the system of equations

Lω = ωz + µ1(z, ω)ωz + µ2(z, ω)ωz + F (z, ω) = 0 (2.3.1)

posed on the whole of the plane (D+ +D− +Γ), whose boundary values on the boundaryΓ satisfy the relation

ω+[α(t)] = (a+ ıb)ω−(t). (2.3.2)

The complex-valued function α = φ + ıψ ∈ C1α(Γ) is such that |α′(t)| 6= 0 everywhere

on Γ; furthermore, α(t) is a bijective mapping of Γ onto itself which preserves the index.With respect to the coefficients µi, F , (a+ ıb), the conditions of Section 2 are assumed tobe fulfilled throughout their domain of dependence. The coefficients of system (2.3.1) aredefined on the complete plane (D+ + D− + Γ) but may have discontinuities of the firstkind on the boundary Γ. We will consider therefore the operator L as two operators

L±ω± = ω±z + µ±1 (z, ω±)ω±z + µ±2 (z, ω±)ω±z + F±(z, ω±) = 0 (2.3.3)

for z ∈ D± correspondingly.Let z± = z±(ζ) be analitic fuctions realizing conformal mappings of the domains

D+1 : |ζ| < 1, D1

− : |ζ| > 1, onto the domains D±, and let t± be their limit values.Then (under the assumption that the loop Γ is a Lyapunov’s curve) problem (2.3.1),(2.3.2) is reduced to the equivalent Riemann’s problem with shift posed in the circle

L±u± = u±z +z±ζz±ζ

µ±1 u±ζ + µ±2 u

±ζ

+(z±ζ)−1

F±(z, u±) = 0,

u+[β(τ)] = (a+ ıb)u−(τ), τ = eıγ, γ ∈ [0, 2π],

where

t+[β(τ)] = α[t−(z)], a+ ıb = a[t−(τ)] + ıb[t−(τ)],

u+(ζ) = ω+[z+(ζ)], u−(ζ) = ω−[z−(ζ)].(2.3.4)

¿From now on we assume therefore that problem (2.3.1) , (2.3.2) is formulated in thecircle |z| < 1 and keep all the previous notation.

In the circle |z| < 1, (z = x + ıy = reıγ), let us consider a function σ = σ(z) =r(φ + ıψ) =

√x2 + y2(φ + ıψ) = σ(r, γ). It is easy to see that the function σ = σ(z)


realizes a homeomorphic mapping of the unit circle onto itself wherein the boundarypoints are connected by the formula α = α(t) ≡ α(eıγ), and that σ satisfies the followingBeltrami’s equation

σz − µ(z)σz = 0,

where

µ(z) = eıγφ− ψ′ + ı(ψ + φ′)

φ− ψ′ + ı(ψ − φ′); |µ|2 =

1 + φ′2 + ψ′2 − 2(φψ′ − φ′ψ)

1 + φ′2 + ψ′2 + 2(φψ′ − φ′ψ),

moreover, by virtue of properties of the functions φ, ψ we have J = (φψ′ − ψφ′) > 0.Indeed, let J = 0 at some point. The differentiation in t of the equality φ2 +ψ2 = 1 leadsthen to the homogeneous system of equations

ψ′φ− ψ′φ = 0, ψ′ψ + φ′φ = 0 (2.3.5)

and, hence, at that point φ′ = ψ′ = 0, which contradicts the assumption |α′(t)| 6= 0.Thus, |µ(z)| ≤ µ0 < 1.

Introducing the new functions ω+(z) = ω[σ(z)], u−(z) = ω−(z), we arrive at thefollowing equivalent Riemann’s problem without shift:

L±u± = u±z + µ±1 u±z + µ±1 u

±z + µ±2 u

±z + F±(z, u±) = 0, (2.3.6)

where µ−i = µ−i , (i = 1, 2), F− = F−, and

µ+1 = −eıθ

(eıθµ− µ1

) (1− eıφµ+

1 µ)

+ µ+2 |µ+

2 |e2ıθ∣∣∣1− eıθµµ+1

∣∣∣2 − ∣∣∣µµ+2

∣∣∣2 ; eıθ =σzσz

; eıφ =σzσz,

µ+2 = −eıθ

(eıθµ− µ

)µµ+

2 −(1− µ+

1 µeıφ)

∣∣∣1− eıθµµ+1

∣∣∣2 − ∣∣∣µµ+2

∣∣∣2 µ+2 ; (2.3.7)

F+ =F+∣∣∣1− eıθµµ+1

∣∣∣2 − ∣∣∣µµ+2

∣∣∣2 ;

Besides, |µ±1 |+ |µ±2 | ≤ µ0(µ±1 , µ

±2 , µ) < 1.

Let us take into consideration the function u∗ = u∗(z), defined for z ∈ D+ in thefollowing way (see [13] ):

u∗(z) = u−(

1

z

).


It is obvious that u∗(t)u−(t) on the boundary of the circle. For the two-component vectorw(z) = u+(z), u∗(z), defined for z ∈ D+, we then get the following Gilbert’s problem:

wz + q1(z, w)wz + q2(z, w)wz +B(z, w) = 0, (2.3.8)

<eΩ(t)w(t) = 0, t = eıγ, γ ∈ [0, 2π], (2.3.9)

where the diagonal matrixes qi, ω, B have the form

qi =

∣∣∣∣∣∣µ+i 0

0 − z2

z2µ−1 ( 1z,w)

∣∣∣∣∣∣ , Ω(t) =

∣∣∣∣∣ 1 −(a− ıb)ı ı(a− ıb),

∣∣∣∣∣B =

∣∣∣∣∣ F+ 0

0 − 1z2F−

∣∣∣∣∣ .To find a solution of problem (2.3.9) in the class of analitic functions, let us consider

the Riemann problem for a single analitic function

φ+(t) = (a+ ıb)Φ−(t). (2.3.10)

All the solutions of problem (2.3.10) from the set of analitic functions bounded at allpoints of descontinuity [23, 24] can be represented in the form

Φ(z) =

Pn(z)X(z) = Φ+(z), z ∈ D+,Pn(z)X(z)z−κ = Φ−(z), z ∈ D−,

(2.3.11)

where Pn is a polinomial of the power not higher than κ with arbitrary coefficients, andX = X(z) is defined by formulas (2.4), (2.7) of Chapter 1. The analitic vector-function

Ψ =Φ+(z),Φ−

(1z

)is then a solution of the corresponding problem (2.3.9). Given

two linearly independent solutions of problem (2.3.9), they correspond to Pn = zκ/2 andPn = ızκ/2. Taking into account formulas (2.4), (2.7) of Chapter 1, sec.2, subsec.2 for X,we construct the canonical matrix of problem (2.3.9):

X(z) = X(z)Z(z) =

∣∣∣∣∣ X(z)zκ/2 ıX(z)zκ/2

eıαX)(z)zκ/2 −ıe−ıαX(z)zκ/2

∣∣∣∣∣=

∣∣∣∣∣ X(z) ıX(z)X(z) −ıX(z)

∣∣∣∣∣×∣∣∣∣∣ zκ/2 0

0 zκ/2

∣∣∣∣∣ .A change analogous to that of Sec. 2, Chapter 1 reduces the problem to the case X(z) =Πnk=1(z − tk)

αkı, so that we will assume this change to be already performed. We seek asolution to problem (2.3.8)-(2.3.9) in the form


w = −0

X (z)

π

∫ ∫K

0

X (z)−1

t− z+ Z(z, t)

0

X (z)−1f(t)

1− tz

dKt = Tκf,

where Z(z, t) =

∣∣∣∣∣ zκ+1 00 zκ+1

∣∣∣∣∣ for κ ≥ 0 and Z(z, t) =

∣∣∣∣∣ t|κ|+1

0

0 t|κ|+1

∣∣∣∣∣ for κ < 0, and

0

X (t)−1 is the matrix inverse to0

X (t). For each f ∈ Lp the function W = Tκf , (κ ≥ 0),satisfies the homogeneous boundary condition and the additional conditions

∫Γ

[0

X (z)]−1

W (z)z−kdθ = 0, z = eıθ, (k = 0, . . . , κ), (2.3.12)

whence we deduce, taking into account (1.1.8), that every component of the vector Wpossesses not less than 2κ+ 2 entire zeroes at the boundary. W = Tκf , (κ < 0), satisfiesthe following boundary condition

<eΩ(t)W (t) = Ω(t)X(t)Re

∣∣∣∣∣∑κ/2−1m=1 (l′m + ıh′m)zm + l′0∑κ/2−1m=1 ((2

m+ıh2m)zm + l20

∣∣∣∣∣ ,where lim, (i = 1, 2) are defined by formulas (1.4) of CHapter 2. Hence, W = Tκf solvesproblem (2.3.9) only under the assumption that κ − 1 complex-valued conditions arefulfilled, following from equality lim, h1

m, (i = 1, 2) to zero.Substituting W = Tκf into equation (2.3.8), we arrive at the nonlinear integral equa-

tion for f = f1, f2

f + Pκf = f + q1Sκf + q2Sκf + BTκf = g, (2.3.13)

where

Sκf =∂Tκf

∂ζ, B =

∣∣∣∣∣∣F+(z,u+)−F+(z,0)

u+(z)0

0 F−(z,u∗)−F−(z,0)u∗(z)

∣∣∣∣∣∣ for u+ 6= 0, u∗ 6= 0,

and B = 0 for u+ = u∗ = 0; g = −B(z, 0).

2. Let us estimate the Lp-norms of the solution to equation (2.3.13), p > 2. Let usfirst show that the linear equation

f + q1Sκf + q2Sκf +B(z)Tκf = g(z), (2.3.14)

where qi =

∣∣∣∣∣ q1j 00 q2

j

∣∣∣∣∣, |qi1|+ |qi2| ≤ q0 < 1, (i = 1, 2), and B is a diagonal matrix with the

coefficients summable with the power p, p > 2, for g = 0 admits only the trivial solution.


Let κ ≥ 0 and let f = f(z) be a solution to equation (2.3.14). Then W = Tκf = W1,W2is a solution of the following boundary problem

Wz + µWz +BW = 0,

<e[Ω(t)W (t)] = 0, (2.3.15)

∫ [0

X

]−1

Wz−kdθ = 0, z = eıθ, (k = 0, 2κ),

where

µ =

∣∣∣∣∣∣ q11 + q1

2(W 1)z

(W1)z0

0 q21 + q2

2(W 2)z

(W2)z

∣∣∣∣∣∣ ,and the function

ω(z) =

W1(z) z ∈ D+,

W2

(1z

)z ∈ D−

is a solution of the homogeneous problem (2.3.5) and the linear equation

ωz + µEωz +BEω = 0, (2.3.16)

where

µE =

µ(z) z ∈ D+,

− z2

z2µ(

1z

)z ∈ D−,

, BE =

B(z) z ∈ D+,

−z2B(

1z

)z ∈ D−,

are defined on the whole of the plane like the coefficients of equation (2.3.1). Accordingto [13], there exists a complete homeomorphism ζ = ζ(z) of the plane z onto the plane ζwhich satisfies the equation

ζz + µEζz = 0 (2.3.17)

such that ζ = ζ(z) ∈ C(p−2)/p(E). Performing the transformation ζ = ζ(z) and returningto W = W1,W2, we get

Wz + B∗W = 0, (2.3.18)

where B∗ = Bζz(1− |µE|2)−1. Moreover, according to estimate of Lemma 2, Section 2,Ch1, B∗ ⊂ Lp′ , p

′ > 2. ¿From here we get the representation for W ,


W (ζ) = H(ζ)Φ(ζ) = e−TκB∗Φ(ζ) = e−TκB∗

∣∣∣∣∣ Φ1

Φ2

∣∣∣∣∣ , (2.3.19)

where Φi(ζ) are analitic on the domain ζ(K), ∈ W 1p (ζ(K)), and =mT∗ρ|Γζ

= 0.

For κ ≥ 0, the vector Φ(ζ) admits the following representation in the class of vector-valued functions at all points of discontinuity

Φ(ζ) = Xζ(ζ)

∣∣∣∣∣ P2k(ζ)Q2k(ζ)

∣∣∣∣∣ ,whereXζ is the canonical matrix of the homogeneous problem, P2k andQ2k are polinomialsof power 2k, real-valued at the boundary. Since BΦ(ζ) satisfies relations (2.3.12) as asolution to problem (2.3.15), either of the polinomials P2k, Q2k must have not less than2k + 2 zeros at the boundary, i.e. P2k ≡ P2k ≡ 0. Therefore, ω ≡ 0 for κ ≥ 0. Thefurther estimating of the solution, as well as the study of the case κ < 0 , are performedcompletely analogous to Sec.2 of Chapter 1. Once the estimate for the solutions is derived,the existence of at least one solution follows the the Schauder Theorem.

Returning now to problem (2.3.1), (2.3.2), let us formulate the theorem.

Theorem 2.3.1 For κ ≥ 0 problem (2.3.1), (2.3.2) is solvable without any extra conditionand its solution contains κ arbitrary constants, for κ < 0 problem (2.3.1), (2.3.2) has aunique solution only if κ− 1 solvability conditions are fulfilled.

3. An analogous method applies to study a more complicated conjugation problemwith shift for equation (2.3.1), whose solution satisfies on the boundary the followingrelation:

a+1 (t)ω+(t) + a+

2 (t)ω+[α+(t)] + a−1 ω−(t)a−2 ω

−[α−(t)] = h(t), (2.3.20)

where µ2, F (z, ω) will be assumed to be equal zero, a±i ∈ Cα, (i = 1, 2), for all t ∈ Γ.We were unable to study exhaustively the questions of existence and the number of

solutions of problem (2.3.1), (2.3.20). We bound ourselves to indicating the scheme ofreduction of this problem to nonlinear integral equations which are equivalent to equationswith completely continuous kernels.

Since all the transformations to be performed in the domains D+ and D− are com-pletely analogous, from now on we work with the function W+(z) and drop the index” + ”. The following notations are used:

f = f(z), fα = f [z(α)], ∂zαu=

∂u[α(z)]

∂z,

αut= ut(t)|tα(z) .

We shall assume that α = α(t) ∈ C1α satisfies the Carleman condition i.e. α[α(t)] = t;

although it is sufficient to claim that αn(t) = t (where n is an integer).


Let ai = ai(z) be smooth continuations of the coefficients in the boundary conditioninside D+ such that |ai(z)|, (i = 1, 2) are harmonic functions, and let α = α(z) be aquasiconformal mapping of D+ onto inself, with a prescribed correspondence α = α(t) ofthe boundary points. Let us perform the following change

a1W + a2

α

W= 0. (2.3.21)

Since α = α(t) is involutive, then α[α(z)] = z. Thus, substituting into (2.3.21) α as theargument, we get

aα1α

W +aα2W = uα. (2.3.22)

To provide solvability of system (2.3.21), (2.3.22) with respect to W andα

W everywherein D+ suffices to claim that at the boundary the following condition holds

||a1(t)| − |a2(t)|| ≥ a0 > 0 for t ∈ Γ. (2.3.23)

Indeed, let (2.3.23) hold at the boundary. Then, taking into account that |ai| are harmonicfunctions, we get

|J(z)| = |a1|α(z)a1(z)− a2(z)aα2 (z)| ≥ ||aα1a1| − |a2|αa2|

= ||aα1a1| − |aα1a2|+ |aα1a2| − |aα2a2||= ||aα1 ||a1| − |a2|+ |a2||aα1 | − |aα2 ||≥ |aα1 |+ |a2|a0 > 0.

Thus,

W =aα1u− a2u

α

aα1a1 − a2aα2= Au−Buα, (2.3.24)

α

W= Aαuα −Bαu.

Expressing Wz, Wz

Wz = Azu−Bzuα + Auz −B∂z

αu

Wz = Azu−Bzuα + Auz −B∂z

αu, (2.3.25)

and substituting them into (2.3.1), we arrive at the equation

Auz − B∂zαu −µ1Auz − µ1B∂z

αu

= −Azu−Bzαu −µ1Azu− µ1Bzu

α = F1 (2.3.26)


Let us transform the terms ∂zαu and ∂z

αu in (2.3.26) and exclude

αuz. Then (2.3.26)

rewrites in the following way

uz +µ1 − εεαQ

1− εεαQuz +

εP

1− εεαQ

αuz=

F1aα1 − FαQ

1− εεαQ, (2.3.27)

where

ε = ε(z) =a2(z)

a1(z), Q(z) =

(1 + µ1

αzαz

)αµ1 +

ααzααz

αzααz ,P (z) = (αz + µ1αz)

[(αz + µ1αzαz + µ1αz

)− µα1

].

Moreover, due to the regularity of the function α = α(t), the modulos of the functionsP and Q are bounded by a constant depending only on µ0 and α′. By virtue of theboundedness of P and Q, ε in (2.3.27) can always be chosen so small (|ε| < ε0) that thefollowing inequality holds:

|µ1 − εεαQ|+ |εP | < |1− εεαQ|. (2.3.28)

Hence, if in the boundary condition the coefficients a1(t), a2(t) are such that

maxt∈Γ |a2(t)|mint∈Γ |a1(t)|

< ε0, (2.3.29)

then condition (2.3.28) is fulfilled, and we come to an integral equation of the type (2.3.13)arguing by analogy with the preceding arguments.

2.4 Problems in two-connected domains

1. Let us formulate and study several so-called mixed problems in the two-connecteddomain G : ρ < |ζ| < R. It is requested to find a solution ω = ω(ζ) ∈ W 1

p of equation(2.2.1) (with F (ζ, ω) ≡ F (ζ)) which at the boundary of G satisfies the following boundaryconditions

<eeıπα(z)ω(ζ) = 0,

where

α =

0, ζ = Reıγ, γ ∈ [0, 2π],−1/2, ζ = ρeıγ, γ ∈ [0, 2π],

(2.4.1)

(we set R = 1) in problem I, and


α =

0, ζ = eıγ, γ ∈ [0, 2π],−1/2, ζ = ρeıγ, γ ∈ [π, 2π],0, ζ = ρeıγ, γ ∈ [π, 2π]

(2.4.2)

in problem II.Problems (2.4.1), (2.4.2) are exhaustively studied in the class of analitic functions in

[13]. We shall seek the solution of problems (2.2.1), (2.4.1), (2.2.1), (2.4.2) in the formω = Λif where Λi are the operators introduced in Sec.1 and f is an unknown analiticfunction from the class Lp(G), p > 2. Substituting ω into equation (2.2.1) we come tothe nonlinear integral singular equation

f + µ1(ζ,Λif)Σif + µ2(ζ,Λif)Σif + F (ζ) = 0. (2.4.3)

If f solves equation (2.4.3), then ω = Λif is a solution of the corresponding problemfor equation (2.2.1). Conversely, if ω is an arbitrary solution of problem (2.2.1), (2.4.1)(2.2.1), (2.4.2), it can be represented in the form ω = Λif where f is a solution to equation(2.4.3). Indeed, according to [13, p.50] every function ω = ω(ζ) ∈ W 1

p admits the uniquerepresentation

ω(ζ) = − 1

π

∫ ∫G

∂ω(t)

∂ζ

t− ζsGt + Φ(ζ), (2.4.4)

where Φ is an arbitrary analitic in G function which in our case must be such thatconditions (2.4.1), (2.4.2) are fulfilled. It follows then that representation (2.4.4) coincideswith Λif where f is the solution to equation (2.4.3). This representation is unique sincethe boundary-value problems (2.4.1), (2.4.2) have unique solutions in the class of analiticfunctions ([13, pp.269-272]).

2. It follows from the properties of operators Λi studi ed in Sec.1.1 that operators Σi

are linear ond bounded in Lp(G), p > 2, and that ‖Σi‖L2 = 1. By continuity of ‖Σi‖Lp

with respect to p, there exists some p, close enough to 2, such that∥∥∥µ1Σif + µ2Σif∥∥∥Lp≤ ‖Σi‖Lp‖f‖Lp = K‖f‖Lp , (2.4.5)

where K = µ0‖Σi‖Lp < 1. Hence, to equation (2.4.5) the principle of contraction mappingis applicable, and the following theorem is true.

Theorem 2.4.1 Equation (2.4.3) and, thereby, problems (2.2.1), (2.4.1), (2.2.1), (2.4.2)(for F (ζ, ω) = F (ζ)) have unique solutions.

Chapter 3

Free Boundary Problems of GasDynamics

3.1 Transformation of equations of gas dynamics

1. The steady, plane, potential flows of a gas are described by the following system of thegas dynamics equations ([33]) :

ρ(q)φx = ψy,ρ(q)φy = −ψx,

(3.1.1)

where q = φ2z+φ

2y is the flow velocity, ρ = ρ(q) is the density, and φ, ψ are, correspondingly,

the potential and the stream function of the flow.The flow is called subsonic if throughout the stream domain the inequality

1

ρ

d(qρ)

dq= 1−M2 ≥ β > 0 (3.1.2)

is fulfilled, where M is the Mach number.Let us denote by qcr the value of velocity when M = 1. Then the flow is subsonic if

q < qcr. It is easy to see that condition (3.1.2) is also the condition of the ellipticity inthe sense of Petrovskii of system (3.1.1).

By virtue of nonlinearity of system (3.1.1), the fulfillment of condition (3.1.2) or, equiv-alently, of the condition max q(z) < qcr, z ∈ Dz, can only be checked when the solution isalready found. It is convenient therefore to use both in theoretical and numerical studyof boundary-value problems for system (3.1.1) the relations between the density and thevelocity which possess the following properties. The density is given by the true relationρ = ρ(q) in a prescribed interval 0 < q ≤ qcr − ε, while for q > qcr − ε it is such that theflow is subsonic and uniform. If the density is given by the relation ρε = ρε(q), the flow iscalled unvariably subsonic. Let us note that under the condition φ2

z+φ2y1/2 ≤ qcr−ε the

solution of the system for the potential obtained for the above-described relation between

49

50 Free Boundary Problems of Gas Dynamics

the density and the velocity will also be a solution of the true equation for the potentialfunction.

¿From now on we assume that condition (3.1.2) is fulfilled for the flow density ρ = ρ(q)(for every q) and that the function ρ = ρ(q) has the derivative which satisfies the Holdercondition, e.g. ρ = ρ(q) ∈ C1

α, α ∈ (0, 1).

2. The principal properties of quasiconformal mappings. Let W = W (z) be a contin-uous differentiable homeomorphism of a plane domain Dz with a positive Jacobian. Sincethe mapping W = W (z) is, evidently, affine, then an infinitely small circle centered at apoint z ∈ Dz transforms into an infinitely small ellips

The ratio Q(z) of the large and small semi-axis of the ellips is the measure of deviationof this mapping from the conformal one. Let us find out Q(z). We have

∂W

∂s=∂W

∂xcos θ +

∂W

∂ycos

(π

2− θ

),

whence

Q(z) =maxθ∈[0,2π] |cos θWx + sin θWy|minθ∈[0,2π] |cos θWx + sin θWy|

.

This relation can be given another form. Denote Wz = 12(Wx− ıWy), Wz = 1

2(Wx + ıWy),

then dW = Wzdz +Wzdz and |dz|||Wz| − |Wz|| ≤ |dz||||Wz|+ |Wz||, whence

Q(z) =max ∂W

∂s

min ∂W∂s

=|Wz|+ |Wz||wz| − |Wz|

=1 +K

1−K

with K = |Wz ||Wz | . If the Jacobian J = |Wz|2 − |Wz|2 ≥ 0, then K ≤ 1.

Free Boundary Problems of Gas Dynamics 51

A mapping W (z) is called Q-quasiconformal (or simply Q-conformal) if Q is uniformlybounded in the domain Dz, i.e., if there exists a number K < such that |Wz| ≤ K|Wz|.The last inequality yields, in particular, that every Q-quasiconformal mapping satisfiesBeltrami’s equation

Wz − µ(z)Wz = 0,

where |µ(z)| ≤ K < 1. The coefficient µ can depend, generally speaking, on W and on itsderivatives, but the mapping will be quasiconformal if, given a solution, |µ(z,W,Wz,Wz)| ≤K < 1 throughout the domain Dz. The solutions of Beltrami’s equation possess manytopological properties intrinsic for conformal mappings. An arbitrary generalized solutionof the Beltrami equation (if the derivatives are understood in the sense of Sobolev) canbe represented in the form

W = Φ[χ(z)],

where χ = χ(z) is the homeomorphism of the Beltrami equation mapping Dz onto thecircle |χ| < 1 and satisfying the same equation

χz − µ(z)χz = 0;

Φ(χ) is a function analitic in the circle |χ| < 1. Let us indicate another property ofanalitic functions proved by B.V.Boyarskii.

Theorem 3.1.1 Let µ(z) be a measurable function satisfying the conditions1. |µ(z)| ≤ µ0 < 1;2. µ ≡ 0 out of some circle centered at the origin.Then there exists a function f(t) such that the function

W (z) = z − 1

π

∫ ∫K

f(t)

t− zdKt

satisfies the Beltrami equation Wz − µ(z)Wz = 0 and realizes a homeomorphic mappingof the plane z onto the plane W , besides, W (z), z(W ) ∈ Cα(E), (0 < α < 1), whereα = α(K,µ0), and E is the plane.

It is proven by N.I.Vekua [13] that if µ(z) ∈ Cmα , then W (z) ∈ Cm+1

α in D, (m ≥ 0).This generalized Riemann’s theorem holds for the quasilinear equations

wz − µ1(z, w)wz − µ2(z, w)wz = 0

in the case where |µ1| + |µ2 ≤ µ0 < 1 for z ∈ K, w ∈ Dw (see [13]). It was provedin papers [10, 34] that under the conditions of normalization, usual in the theory ofconformal mappings, the solution of a quasilinear equation mapping a circle onto anarbitrary domain, is unique.

Theorem 3.1.1 can be given the following generalization.


Theorem 3.1.2 If µ(z) is a function measurable on the whole of the plane and the con-dition |µ(z) ≤ µ0 < 1 holds, then there exists a complete homeomorphism W = W (z) ofBeltrami’s equation which belongs to the class Cα(E).

If µ(z) ∈ Cmα , then there exists a homeomorphism W (z) ∈ Cm+1

α .

Theorem 3.1.3 Let W n(z) be a sequence of homeomorphisms of the unit circle K ontoitself. Assume that each W n satisfies the equation

W nz − µn(z)W

nz = 0

and the norming condition W n(0) = 0, W n(eıγn) = eıγ0. Let µn → µ everywhere in Kand |µn(z)| ≤ µ = 0 < 1 for all n. Then the sequence W n(z) converges uniformly, thelimit W (z) = limn→∞W n(z) satisfies the equation

Wz − µ(z)Wz = 0

and is a homeomorphic mapping of the circle onto itself which preserves the normingconditions.

(see [10]). This theorem is true for the homeomorphisms of an arbitrary domain ontoitself.

3. The steady invariable subsonic potential flow of a gas is described by the followingsystem of the gas dynamics equations

ρ(q)φx = ψy,ρ(q)φy = −ψx, q2 = φ2 + ψ2,

(3.1.3)

besides,

1

ρ

d(ρq)

dq= 1−M2 ≥ β > 0 for all q. (3.1.4)

Condition (3.1.4) provides the uniform ellipticity in the sense of Petrovskii of system(3.1.3). Introducing, analogous to [33], the fictitious density and velocity

q∗ =∫ q

1

√1−M2

qdq, ρ∗ =

ρ(q)√1−M2

,

(it is easy to see that q∗ has a logarithmic singularity when q = 0 and q = ∞), let ustransform the well-known system of equations of godograph (see [33])

φθ = qρψq,

φq = −1−M2

qρψθ,

(3.1.5)

to the form


ρ∗(q∗)φθ = ψq∗ρ∗(q∗)φq∗ = −ψθ,

(3.1.6)

or, for the inverse functions q∗ = q∗(φ, ψ) and θ = θ(φ, ψ),ρ∗q∗ψ = θφ,ρ∗θψ = −q∗φ.

(3.1.7)

Passing to the complex-valued functions w = φ + ıψ, ω = q∗ − ıθ, we can rewritesystem (3.1.7) as follows:

ωw −ρ∗ − 1

ρ∗ + 1ωw = 0. (3.1.8)

For the incompressible liquid q∗ = ln q, ρ∗ = ρ = 1, and ω = ln dwdz

is an analiticfunction. Transforming system (3.1.3) to the inverse functions and passing to the complex-valued functions, we arrive at the following system of equations for the function z = z(w):

zw −ρ− 1

ρ+ 1zw = 0. (3.1.9)

Substituting into equation (3.1.8) the expressions for ωw and ωw,

ωw = ωzzw + ωzzw,ωw = ωzzw + ωzzw,

we have

ωz −zwzw

ρ−1ρ∗+1

− zw

zw(1− ρ∗−1

ρ∗+1zw

zw

)ωz = 0.

Taking into account (3.1.9) and the relation zw

zw= e2ıθ, we come to the equation

ωz − e2ıθ1−

√1−M2

1 +√

1−M2ωz = 0. (3.1.10)

The invariable character of the subsonic flow implies the following estimates, uniform withrespect to q:

∣∣∣∣∣ρ∗ − 1

ρ∗ + 1

∣∣∣∣∣ ≤ µ0 < 1,

∣∣∣∣∣ρ− 1

ρ+ 1

∣∣∣∣∣ ≤ µ0 < 1,

∣∣∣∣∣e2ıθ 1−√

1−M2

1 +√

1−M2

∣∣∣∣∣ ≤ µ0 < 1. (3.1.11)

These estimate provide the uniform eelipticity of the corresponding systems. Let usintroduce the new independent complex variable τ = x+ ıψ and rewrite system (3.1.8) in


terms of τ . Substituting into (3.1.8) the expressions ωw = ωττw+ωττw, ωw = ωττw+ωττw,we get, taking into account (3.1.3), (3.1.9):

ωτ − µ(ω)ωτ = 0, (3.1.12)

where

µ(ω) =ρq2 − u

√1−M2 + ıθ

ρq2 + u√

1−M2 − ıθ, u = φx, v = φy. (3.1.13)

In the incompressible liquid µ(ω) = eω−1eω+1

. The system of equations corresponding toequation (3.1.12) is elliptic but the degeneracy is admitted at the points where |µ| = 1.To make the mapping z = z(τ) homeomorphic and, hence, to make it possible to pass tothe variable τ , it suffices to fulfill the condition

|µ(ω)| ≤ µ0 < 1. (3.1.14)

Condition (3.1.14) is fulfilled if throughout the flow domain the following inequality holds

| ln q| ≤M <∞, |θ| < π

2− ε0, ε0 > 0.

For the sought solution ω = q∗ − ıθ the last conditions take on the form

|<e ω| ≤ N <∞, |=mω| < π

2− ε0, ε0 > 0. (3.1.15)

These conditions are the conditions of ellipticity of system (3.1.12) at the solution ω =ω(τ). To fulfill conditions (3.1.15) throughout the domain Dz it suffices to claim thatthey are fulfilled at the boundary Γz of Dz. Let us continue µ to the whole of the planein the following way:

µ =

µ(ω), z ∈ Dz,0, z 6∈ Dz.

The function ω = ω(z) can be represented then in the form ω(z) = Φ[χ(z)], whereχ = χ(z) is a homeomorphism of the whole of the plane onto the whole of the plane, andΦ(χ) is a function analitic on Dχ (on the image of Dz). If conditions (3.1.15) are fulfilledon Γz then

|<eΦ(χ)| ≤ N <∞,

|=mΦ(χ)| < π

2− ε0, ε0 > 0 on Γχ.

By the maximum principle for harmonic functions it follows then that the last inequalitiesare fulfilled everywhere in Dχ, whence their validity in Dz. Thus, if the sought solution


satisfies conditions (3.1.15) everywhere on the boundary, then system (3.1.12) is uniformlyelliptic everywhere in the domain Dτ (the image of Dz) and the transformation to thevariable τ is therefore justified.

The mapping of the corresponding domains is given by the following schemes:

3.2 Statement of problems of gas dynamics

1. Free-boundary problems in two-connected domains. We consider Problem (A) aboutthe flow of an incompressible fluid in a channel with completely prescribed (or completelyunknown) walls near an unknown (or a prescribed obstacle).


It is assumed that on the unknown (free) parts of the boundary of the flow domain (onthe channel walls or on the unknown obstacle, which is drawn with the dotted line) thevelocity distribution is given: q = q(x) ∈ C1

α, besides, q = q(x) 6= 0 and q(±∞) = 0. Onthe prescribed parts of the boundary we have θ = θ(x) ∈ C1

α; furthermore, |θ| < π2− ε0,

ε0 > 0 and θ(±) = 0 in the problem where the shape of the channel walls is unknown. Inthe problem of streamlining of an unknown obstacle, the points O and O′ are assumedto be the branch point and the point of getting off the stream, therefore θ = 0 at thesepoints. Throughout the boundary Lz we have ψ = const (ψ = 0 on the obstacle andψ = ±π

2on the channlel walls).

It is requested to define tha form of the unknown portions of the boundary of the flowdomain Dz, which need not to be univalent, and the distribution of the univalent complexvelocity q(z)e−ıθ(z) in the flow domain.

2. Let us consider Problem (B) of the cavitational streamlining of a system of non-symmetric curvelinear arcs by a potential uniformly subsonic gas jet via a scheme of theRyabushinskii type (see fig.3)

It is assumed that the arcs AOB and COD are connected by a partition OO whoselength (and, thus, l) are not prescribed, and that in the cavities OOBC and OAO thedifferent constant pressures P1 and P2 are maintained. According to Bernulli’s law, theycorrespond to constant velocities q1 , q2 on the free boundaries BC and A (the valuesof the parameters will be defined later on). Let the points O and O be the points ofbranching and getting off the stream where the angle of inclination of the velocity O = 0,and the value of the velocity is bounded at the points A, B, C. Along the prescribed arcsBOA and CO′D the function θ(x) = arctan dy

dx∈ C1

α is given, besides, |θ| < π2− ε0 . The

stream function ψ = 0 along the whole of the boundary of the flow domain Dz.


3. Let us define the images Dτ of the domains Dz for each of the problem formulated,and formulate the boundary-value problems for the function ω = ω(τ) on the plane Dτ .

For Problem (A), the domain Dτ is the strip −π2≤ ψ ≤ π

2, cut along the real axis (see

fig.4)

For problem (B), the domain Dτ is the plane cut along the axis Ox, x ∈ [0, l] (seefig.5)

4. It is known from the general theory of conformal mappings that every two-connecteddomain can be maped via a conformal mapping onto a ring bounded by two coaxialcircumferences. The ratio of the radius of the circumferences is defined by the geometricshape of the domain.

In each of Problems (A) (Problem (B) will be considered separately), let us map the


domain Dτ in a conformal way onto the ring ρ < |ζ| < 1 so that the image of theunknown part L2

z is the circumference |ζ| = 1, and the image of the prescribed part L′z isthe circumference |ζ| = ρ. Thereby a one-to-one correspondence between the boundariesLτ and Lζ is established. Let ζ = reıγ, then the correspondence x = x(γ) is given, andfor the function ω = ω(ζ) we get the following boundary conditions:

<e ω[eıγ] = q∗[x(γ)], γ ∈ [0, 2π],=mω[ρeıγ] = −θ[x(γ)], γ ∈ [0, 2π].

(3.2.16)

On the plane ζ, equation (3.1.12) is reduced to the form

ωζ − µωζ = 0, (3.2.17)

where µ = µ exp−2ı arctan dτ

dζ

, and τ = τ(ζ) is an analitic function mapping the ring

onto the corresponding domain Dτ . Obviously, under this transformation |µ| = |µ|.

3.3 A priori estimates of the solution

1. Let us show that if q = q(x) ∈ C1α, q 6= 0 along the unknown parts of the boundary,

and θ = θ(x) ∈ C1α on the prescrined parts, with |θ| < π

2− ε0, ε0 > 0, then throughout

the flow domain the sought solution satisfies conditions (3.1.15) of Section 2.Let us study first the regularity of the sought solution under the assumption that it is

bounded. To begin with, let us assume the fluid to be incompressible, e.g., ω(w) = ln dwdz

and there exists a bounded solution, ω = − ln(xψ + ıyψ); then, evidently, z = z(w) ∈Cγ, 0 < γ ≤ 1. Furthermore, for the analitic function ω = ω(w[z(w)] we get theboundary-value problem (3.2.16) posed on the plane w, with ln q[x(φ)], θ = θ[x(φ)]Holder-continuous functions of the variable φ. But then the corresponding problems areuniquely solvable and ω = ω(w) ∈ Cγ in Dw and, hence, z = z(w) ∈ C1

γ . Correspond-ingly, ln q[x(φ)], θ = θ[x(φ)] possess the first derivatives which are Holder-continuous.Analogous to preceding arguments, it follows from the representation for the solution ofthe corresponding problem that in Problems (A) ω(w) ∈ C1

γ(Dw).

2. Let us prove that if q = q(x) ≡ const on L2z, the second of conditions (3.1.15) is

fulfilled in Dz. Under the above condition the harmonic function θ(z) = =m dwdz6≡ const

cannot attain its absolute extremum on L2z. Indeed, at the point of the absolute extremum

of a harmonic function θ the inequality∣∣∣ ∂ θ∂ n

∣∣∣ > 0 must be fulfilled but, due to the Cauchy-Riemann conditions, ∣∣∣∣∣∂ θ∂ n

∣∣∣∣∣ =∣∣∣∣∣∂ ln q

∂s

∣∣∣∣∣ = 0 on L2z.

Thus, if |θ| < π2−ε0 on L′z, then former inequality holds throughout Lz and, henceforth,

everywhere on Dz.


3. Let us prove the first of the inequalities (3.1.15) for the sought solution of Problems(A) under the assumption that q ≡ const = 1.

Let us cut the domain Dz by the streamline ψ = 0; according to what was provenabove, |θ| < π

2− ε0 along all this streamline. Moreover, the cut may correspond to two

non-coinciding similar curves, which means non-univalence of the domain Dz.The solution of our problem is the logarithmic derivative of an analitic function which

realizes a mapping of the domain Dw onto the domain Dz. Let us estimate ln∣∣∣dwdz

∣∣∣ in the

domains D′w and D′′

w.Let us consider first the problem of the flow in a channel with the completely unknown

walls near a prescribed obstacle. Assuming that θ is given everywhere on the line ψ = 0,let us construct a solution ω = ω(w) in D′

w :0 ≤ ψ ≤ π

2

. According to the Ville

formulas for the strip, we have

ω(w) =1

π

∫ ∞

−∞

θ[x(t)]

sinh(t− w)dt+

∫ ∞

−∞

ln q[x(t)]

cosh(t− w)dt =

1

π

∫ ∞

−∞

θ[x(t)]

sinh(t− w)dt (3.3.1)

since q ≡ 1. The integral in (3.3.1) has a continuous limit at the line ψ = π2. Let w → φ,

then the integral can be rewritten in an equivalemt from with the use of the classical trick:

1

π

∫ ∞

−∞

θ[x(t)]− θ[x(φ)]

sinh (t− φ)dt− θ[x(φ)]

1

π

∫ ∞

−∞

dt

sinh (t− φ), (3.3.2)

where

1

π

∫ ∞

−∞

dt

sinh (t− φ)= i when w ∈ D′

w,

while the first term on (3.3.2) is continuous when w → φ due to Holder continuity of θ.Passing to the limit as w → φ, we get that on the real axis

ω(φ) = −ıθ +1

π

∫ ∞

−∞

θ[x(t)]− θ[x(φ)]

sinh (t− φ)dt = −ı θ + =m(φ.) (3.3.3)

According to what we said above, it is necessary to estimate the absolute value of =m(ψ)on that part of the real axis which corresponds to the prescribed obstacle.

Accept the notation H = max∣∣∣ln ∣∣∣dw

dz

∣∣∣∣∣∣ (by virtue of the smoothness assumption it is

necessarily attained), and proceed to estimate |=m(φ)|. Let us rewrite J in the followingway:

J =1

π

∫ φ−η/2

−∞

θ[x(t)]− θ[x(φ)]

sinh (t− φ)dt+

1

π

∫ φ+η/2

φ−η/2

θ[x(t)]− θ[x(φ)]

sinh (t− φ)dt

+1

π

∫ ∞

φ+η/2

θ[x(t)]− θ[x(φ)]

sinh (t− φ)dt = J1 + J2 + J3, (3.3.4)


where φ ∈ [0, ], and η is, for the moment, an arbitrary positive number. We estimate theintegrals Ji in the same way as in [35].

|J2| =

∣∣∣∣∣ 1π∫ φ+η/2

φ−η/2

θ[x(t)]− θ[x(φ)]

sinh (t− φ)dt

∣∣∣∣∣ ≤ max

∣∣∣∣∣dθdφ∣∣∣∣∣ 1

π

∫ φ+η/2

φ−η/2

tdt

sinh t

≤ η

πmax

∣∣∣∣∣dθdφ∣∣∣∣∣ ≤ max

∣∣∣∣∣dxdφ∣∣∣∣∣max

∣∣∣∣∣dθdx∣∣∣∣∣ ≤ η

πeH max |θ′|, (3.3.5)

|J2 + J3| ≤2

π|max |θ|

∣∣∣∣∣∫η/2∞

tdt

sinh t

∣∣∣∣∣ = 2

πmax |θ|

∣∣∣∣ln tanη

4

∣∣∣∣ . (3.3.6)

Taking into account (3.3.5), (3.3.6), we get from (3.3.3) that

H ≤ 2

πmax |θ|

∣∣∣∣ln tanhη

4

∣∣∣∣+ η

πeH max |θ′|. (3.3.7)

Let us set η = e−H , 0 < η < 1. Then∣∣∣∣ln tanhη

4

∣∣∣∣ ≤ | ln η|+∣∣∣∣ln tanh

1

4

∣∣∣∣ (3.3.8)

and (3.3.7) gives

H ≤(1− 2

πmax |θ|

)−1 ( 2

πmax |θ|

∣∣∣∣ln tanh1

4

∣∣∣∣+ 1

4max |θ′|

), (3.3.9)

or, if the inequalities |θ| < π2− ε0, ε0 > 0, are taken into account,

H ≤ ε−10

(2

πmax |θ|

∣∣∣∣ln tanh1

4

∣∣∣∣+ 1

4max |θ′|

)= N1

(ε−10 , max |θ|, max

∣∣∣∣∣dθdx∣∣∣∣∣).

Estimating in D′′w is performed in the completely analogous way. Choosing N =

max (N1, N2), (i = 1, 2 correspond to the domains D′w, D′′

w), we have that throughout theflow domain

|<e ω| ≤ N

(ε−10 , max |θ|, max

∣∣∣∣∣dθdx∣∣∣∣∣). (3.3.10)

Therefore, we have proved the fulfillment of conditions (3.1.15) for the thought solutionof the problem of the flow in the channel with the completely unknown walls.

For the problem of the flow in a prescribed channel the estimate of the solution followsin an analogous way via the representation for the solution of the mixed problem posedon a strip.


3.4 The existence theorem

Let us formulate a theorem needed in what follows. ([36])The Leray-Schauder Theorem Let F (f, γ) be an arbitrary one-parameter family

of completely continuous mappings of a Banach space E onto itself, and let λ ∈ [0, 1]. LetG be a bounded domain of the space E with the boundary Γ. Assume that the mappingsF (f, λ) are equicontinuous in λ throughout G, e.g., for every ε > 0 thete exists δ > 0such that if |λ′ − λ′′| < δ, then ‖F (f, λ′)− F (f, λ′′)‖ < ε for all f ∈ G.

Let f − F (f, λ) 6= 0 for f ∈ Γ and

1. the equation f − F (f, λ) = 0 has a unique solution f0 ∈ G.

2. if the mapping g = f − F (f, 0) is locally bijective in some neighborhood of f0, thenthe mapping

f − F (f, λ) = 0

for every λ ∈ [0, 1] has at least one solution in G.

The operators F (f, λ) can be thought of as defined on a bounded set M ⊂ E, butmapping this set again into a set N ⊂ E. By virtue of the well-known theorem ([27])they can be continued then by completely continuous operators to the whole of E.

Theorem 3.4.1 Let F be a completely continuous operator defined on a closed set M,and let N be a convex hull of the set of values which this operator takes on ; let ε > 0 bea given number. Then there exists an operator F it is defined on the whole of the spaceE, coinciding with F on M, and such that ρ[F (f),N ] < ε for every element f ∈ E.

In this section, we shall prove the solvability of those problems of hydrodynamics intwo-connected domains whose solutions are proved to satisfy inequalities (3.1.15) on thewhole of Lz, i.e., for the problems of hydrodynamics of incompressible fluids with q ≡ 1along the unknown boundaries.

Moreover, for the problem of the flow in a prescribed channel we shall additionallyassume that θ = θ(x) and its first derivative decrease fast enough at infinity, so thatθ = θ[x(γ)] ∈ Cα in a neighborhood of the image of the infinity.

In §2 of this Chapter, free-boundary problems of gas dynamics were reduced to thefollowing equivalemt boundary-value problem posed in the ring ρ < |ζ| < 1

<e ω[eıγ] = ln q[x(γ)], γ ∈ [0, 2π],=mω[eıγρ] = −θ[x(γ), γ ∈ [0, 2π]

(3.4.1)

for the equation

ωζ − µ(ζ, ω)ωζ = 0, (3.4.2)


where

µ(ζ, ω) =eω − 1

eω + 1exp

2ı arg

dτ

dζ

,

and τ = τ(ζ) is an analitic function mapping the ring onto a corresponding domain Dτ .Let us consider a one-parameter family of functions ω(ζ, λ) which for every λ ∈ [0, 2π]

satisfy equation (3.4.2) and the following boundary conditions:

<e ω[eıγ] = 0, γ ∈ [0, 2π],=mω[eıγρ] = −λθ[x(γ), γ ∈ [0, 2π].

For λ = 1 the solution of problem (3.4.2)-(3.4.3) coincides with the solution of the cor-responding boundary-value problem (3.4.1)-(3.4.2) (with q ≡ 1). The value λ = 0 cor-responds to the unique solution ω0(ζ) ≡ ω(ζ) ≡ 0. Indeed, every solution of equation(3.4.2) can be represented in the form ω = Φ[χ(ζ)], where χ = χ(ζ) is a homeomorphismof the domain Dζ which satisfies the equation

χζ − µ[ζ, ω(ζ)]χζ = 0,

and Φ = Φ(χ) is a function analitic inDχ (in the image ofDζ), satisfying the homogeneousconditions (3.4.3) (λ = 0). On the other hand, according to [13, p.260], the only solutionadmitted by the homogeneous problem (3.4.3) in the class of analitic functions is thetrivial one, whence ω0 = Φ[χ(ζ)] ≡ 0. The solution ω = ω0(z) ≡ 0 in the plane zcorresponds to the flow with the equation y = const of the known boundaries of L′z, andwith the velocity q ≡ 1 on L2

z.Let us make use of the Leray-Schauder method to prove solvability of problem (3.4.1)-

(3.4.2) for every λ ∈ [0, 1]. We shall prove thereby the solvability of the original problem(3.4.1)-(3.4.2) with λ = 1. As was already shown in §2 of Chapter 2, any solution of theboundary-value problem (3.4.1)-(3.4.2) can be represented in the form

ω(ζ, λ) = Λ1f + λΦ(ζ),

where Φ(ζ) is an analitic function which satisfies the boundary conditions (3.4.1) (q ≡ 1),Λ1 is the operator introduced in §1 of Chapter 2, and f is an unknown function of theclass Lp(K), p > 2, defined from the condition that the function ω − ω(ζλ) satisfiesequation (3.4.2). Substituting ω into equation (3.4.2), we get the following family ofsingular integral equations

f − µ [ζΛ1f + λΦ(ζ)] Σ1f − λdφ

dζµ [ζΛ1f + λΦ(ζ)] (3.4.3)

which are equivalent to problem (3.4.1)-(3.4.2). It follows from results of §3 of Chapter3 that the sought solutions ω(z) = ω[ζ(z)] satisfy on the whole of the domain K theinequalities


|<e ω(ζ, λ)| < N <∞, and |=mω(ζ, λ)| < π

2− ε0, ε0 > 0,

where N and ε0 can be chosen independent of λ. Let us denote by G the set of thosef ∈ Lp(K), p > 2, for which the inequalities

|<e (λΦ + Λ1f)| < N and |=m (λΦ + Λ1f)| < π

2− ε0.

Let us regard all operators in equation (3.4.3) as defined on the closure G of the set G.By virtue of the diffentiability of µ with respect to its second argument, and since theoperator Λ1 is completely continuous, the operator

AΛ1f = λµ (ζ,Λ1f + λΦ)d : Phi

dζ

is completely continuous in Lp, p > 2. It is obvious that for all f ∈ G the inequality|µ(ζΛ1f + λΦ)| ≤ µ0 holds. By the Riesz theorem applied to ‖Σi‖Lp , there is a p, closeenough to 2 , such that

‖µ (ζ,Λ1f + λΦ)‖Lp≤ K‖f‖Lp , where K = µ0‖Σi‖Lp < 1.

Hence, there exists a linear operator (I − µΣ)−1, bounded on the set G, and continuouslydepending on ω = Λ1f+λΦ as a parameter. That is why equation (3.4.3) can be rewrittenin the equivalent form

f = λF (ω), (3.4.4)

where the operator

F (f, λ) = λF (ω) = (I − µΣ1)−1

[λµ(ζ,Λ1f + λΦ)

dΦ

dζ

]

is completely continuous with respect to f ∈ G for every fixed λ, if F continuously dependson ω as a parameter.

Let us show that the operator F (ω) continuously depends on ω. Let ω1 − ω2. Forf = f1 − f2 = λ[F (ω1)− F (ω2)] we have the following equation

f − µ(ζ, ω1)Σ1f − [µ(ζ, ω1)− µ(ζω2)]

(Σ1f2 + λ

dφ

dζ

)= 0

and, since µ is continuous with respect to ω,

‖f‖Lp ≤1

1−K

∥∥∥∥∥Σ1f2 + λdφ

dζ

∥∥∥∥∥Lp

max |µ(ζ, ω1)− µ2(ζ, ω1)|

→ 0 when ω1 → ω2.


Equicontinuity in λ of the operator F (f, λ) = λF (ω) with fixed f follows from the con-tinuity of F (ω) in ω = Tf + λΦ. Since the sought solution ω satisfies the inequalities|<e ω| < M , |=mω| < π

2− ε0, then, obviously, f 6= F (f, λ) for f ∈ Γ ≡ (G − G). For

λ = 0 we have F (f, 0) = λF (Tf + λΦ)∣∣∣λ=0

= 0, hence, f0 = 0 will be the unique solution

of equation (3.4.4). In this case the second condition of the Leray-Schauder theorem takeson the form g = f , and the local bijectivity of the correspondence of g and f becomesobvious.

We have proven therefore the fulfillment of all conditions of the Leray-Schauder the-orem whence the solvability of the boundary-value problem (3.4.1)-(3.4.2) and, hence,the existence of solutions of the original problems of hydrodynamics with q ≡ 1 on theprescribed boundaries.

To establish the existence theorem in the case of an arbitrary q = q(x) ∈ C1α, let us

follow [1, 3] and make use of the parameter extension method. Let us claim first someadditional regularity of q = q(x) in a neighborhood of the points ±∞. To begin with, let

us assume that ln q(x) and d ln q(x)dx

vanish fast enough at infinity so that ln q[x(γ)] ∈ Cαin a neighborhood of the point γ0 which is the image of the infinity (cf. formula (3.2.16)).

Let us consider the one-parameter family of the boundary-value problems

<e ω = λ ln q(x),=mω = −θ(x). (3.4.5)

Let us assume theexistence of a bounded solution ω, then, by analogy with §3 of Chapter3, formula (3.3.1) is true for ω in the strip D′

w. Passing in that formula to the limit asw → φ+ ıπ

2, and separating then the real and imaginary parts of ω, we get

θ = −λπ

∫ ∞

−∞

ln q[x(t)]− ln q[x(φ)]

sinh (t− φ)− 1

π

∫ ∞

∞

θ(t)dt

cosh (t− φ)= J1 + J2. (3.4.6)

Let us estimate the integrals Ji.

|J1| ≤ λ

πmaxL2

z

∣∣∣∣∣d ln q

dx

dx

dφ

∣∣∣∣∣∣∣∣∣∣∫ ∞

−∞

t− φ

sinh (t− φ)dt

∣∣∣∣∣≤ λmax

L2z

∣∣∣∣∣d ln q

dx

∣∣∣∣∣maxL2

z

1

q= λM,

where

M = M

(maxL2

z

∣∣∣∣∣d ln q

dx

∣∣∣∣∣ ,maxL2

z

| ln q|)

is a constant not depending on the solution.


|J1| ≤maxL2

z|θ|

π

∣∣∣∣∣∫ ∞

−∞

dt

cosh (t− φ)

∣∣∣∣∣ ≤ maxL′z

|θ‖.

Thus,

|θ| ≤ maxL′z

|θ|+ λM <π

2− ε0 + λM. (3.4.7)

Let us choose in (3.4.7) ε0 so small that the inequality

ε0 − λ0M > ε1 > 0 (3.4.8)

is true. Given λ = λ0, the inequality |θ| < π2− ε1, ε1 > 0, is fulfilled throughout the

domain D′w. The same estimate is true in D′′

w and, therefore, in the whole of the domainDw we have |θ| < π

2− ε1, ε1 > 0. Making use once again of formula (3.3.1), analogously

to §3 we arrive at the estimate of |<e ω| in the domain Dw:

|<e ω| < N

(ε−11 , max

L′z|θ|, max

L′z| ln q|, max

L′z|d ln q

dx|, max

L′z|θ′|)<∞.

Hence, given λ = λ0, problem has a solution. Let us repeat this procedure for the problem

<e ω = (λ0 + λ1) ln q(x),=mω = −ω(x)

(3.4.9)

with λ1 being chosen by the condition

ε1 − λ1M > ε2 > 0. (3.4.10)

In this continuous extension in the parameter two situations can occur. Either we canalways choose an appropriate λ > 0 and finally arrive at the original problem, or theprocess stops for some λ = λ∗. Assume that the latter happens at the step λ − λ∗ = 0,that is, there exists at least one point z0 on L2

z where |θ| = π/2. If this is the case,the point z0 is a point of extremum of the harmonic function θ and, according to themaximum principle,

∣∣∣ ∂θ∂n

∣∣∣∣∣∣z=z0

> 0 . On the other hand, the Cauchy-Riemann conditions

on L2z imply ∣∣∣∣∣∂θ∂n

∣∣∣∣∣∣∣∣∣∣z=z0

=

∣∣∣∣∣d ln q(x)

dx

∣∣∣∣∣∣∣∣∣∣z=z0

cos θ = 0

i.e. the extremum cannot be attained. Hence, one can realize the extension in theparameter λ until λ = 1. Let us construct the sequence of the auxiliary problems

<e ω = ln qn[x(γ)], |ζ| = 1,=mω = −θn[x(γ)], |ζ| = ρ,ωnζ − µ(ζ, ωn)ωnζ = 0,


where qn[x] → q(x), θn → θ(x) as n → ∞, and for every n qn satisfy an additionalcondition at infinity. Applying Theorem 3 of §1, Chapter 3, analogously to [1, 3] weconclude that the family of solutions ωn is compact whence the existence of solutions toProblems (A) for arbitrary q(x) ∈ C1

α, θ(x)C1α, |θ| < π

2− ε0, ε0 > 0.

3.5 The proof of existence of a solution for a Ryabushinskii-

type problem

1. Let us consider Problem C formulated in §2. For the solution of problem (3.1.12)ω(τ) = q∗ − ıθ we get the boundary-value problem with the boundary conditions posedon the image of the flow domain Dz of the plane τ = x+ ıy (see fig.6)

<e ω(x) = q∗i on BC and AD,=mω(x) = −θi(x) on AOB and CO1.

(3.5.1)

Since the velocity is constant, q = q(x) ≡ const, along the jets BC AD, proceeding byanalogy with works [1, 8] we infer that the absolute extremum of the function θ(z) cannotbe achieved at these jets. That is, θ(x)π

2− ε throughout the boundary of the flow domain

and, due to the maximum principle for θ(x), the same is true in the closed domain Dz.The second of inequalities (3.1.6) for the function ω(τ) is thus proven. Let us prove thefirst of inequalities (3.1.6) for the solution ω(τ) bounded at the points A, B, C, and D.

According to results of §2 (see formula (3.1.9)). the function z = z(w) satisfies theequation

zw −ρ(q)− 1

ρ(q) + 1zw = 0, (3.5.2)

which is a uniformly elliptic equation due to the unvariable character of the subsonic flow.

Assume that the solution was found and substituted into ρ(q). Then the equationrenders Beltrami’s equation. Let us consider an element lz of one of the prescribed arcsand construct a domain dz ⊂ Dz, with a smooth boundary whose diameter is not lessthan some fixed ε, belonging entirely to the interior of Dz and having a common arc lzwith Dz. By virtue of the already proved inequality |θ| < π

2− ε, there always exists such

a domain, which, moreover, does not intersect the other boundaries of the flow domainDz.

Let us map by means of a conformal mapping the domain dw (the image of the domaindz in the plane w = φ+ ıψ) onto the circle |ζ| < 1. Then the function z = z(x), which is aquasiconformal mapping of this circle onto the domain dz, satisfies the Beltrami equation

zζ − µ1(ζ)zζ = 0,


where µ1(ζ) = ρ−1ρ+1

e−2ı arg dwdz and |µ1(ζ)| ≤ a(ρ) < 1. To estimate the quasiconformal

mapping z = z(ζ) of the domain dz onto the circle, let us preliminary map the circle|ζ| < 1 onto the circle |χ| < 1 with the fixed points ζ = 0 and ζ = 1 via the solution ofthe equation

χζ − µ1χ(ζ) = 0.

It follows from the classical results of the theory of quasiconformal mappings that χ(ζ) ∈Cα0 in the closed circle |ζ1 ≤ 1, where 0 < α0 ≤ 1. According to paper [37], for theanalitic function z = z(χ), which realizes a conformal mapping of the circle |χ| < 1 ontothe domain dz, the following unequality is true∣∣∣∣∣ln

∣∣∣∣∣ dzdχ∣∣∣∣∣∣∣∣∣∣ < m(ε) for |χ1 ≤ 1

whence, in particular, |x(χ1)−x(χ2)| ≤ K(ε)|χ1−χ2| for |χ1 ≤ 1, (x = <e z). Observingthat χ(ζ) ∈ Cα0 in the closed circle |ζ| ≤ 1, we get: |x(ζ1)− x(ζ2)| ≤ K1(ε)|ζ1 − ζ2|α0 for|ζi| ≤ 1. Hence, under the quasiconformal mapping dz onto the circle |ζ| < 1

‖θ[x(γ)]‖Hα0= sup

γ1, γ2

|θ[x(γ1)]− θ[x(γ2)||γ1 − γ2|α0

≤ supγ1, γ2

|θ[x(γ1)]− θ[x(γ2)||x(γ1)− x(γ2)|

K1(ε)

and

‖θ[x(γ)]‖Cα0≤ K2

(ε,

∥∥∥∥∥dθdx∥∥∥∥∥C

),

where γ ∈ l is the image of lz on the circumference | | = 1. Since the inequality obtaineddoes not depends on lz, for the sought quasiconformal mapping of the domain Dz ontothe circle | | < 1 we have that

‖θ[x(γ)]‖Cα0≤ K2

(ε,

∥∥∥∥∥dθdx∥∥∥∥∥C

)

along the images of the arcs AOB and CO1D on the circumference | | < 1.Let us consider the equation for the function ω = ω(z) (see §2)

ωz − e2ıθ1−

√1−M2

1 +√

1−M2= 0.

Let the solution of the original problem be found. Then, substituting this solutioninto the coefficient of the last equation, for ω(z) we get the Beltrami equation

ωz − µ(z)ωz = 0,


where |µ(z)| ≤ µ0 < 1 due to the unvariable character of the subsonic flow. Let us mapconformly the domain Dz onto the circle |ζ1 < 1, with points A, B corresponding to thepoints 1, ı, −1, by means of the solution of Beltrami’s equation

ζz − µ(z)ζz = 0.

For the analitic function ω(ζ) we get then the boundary-value problem

<e ω(γ) = q∗i , γ ∈[

π2, π]+ [γ0, 2π]

=mω(γ) = −θj[x(γ)], γ ∈

[0, π

2

]+ [π, γ0]

.

(3.5.3)

Moreover, by virtue of what was proved above, ‖θ[x(γ)]‖Cα0≤ K2. The solution of the

last problem bounded at the points ±1, ı, and eıγ0 can be represented in the form

ω(z) =R(ζ)

2π

∫ 2π

0

h(γ)

R (eıγ)

eıγ + ζ

eıγ − ζdγ,

where R(ζ) =√

(ζ2 − 1)(ζ − ı)(ζ − eıγ0)ζ−1,

h(γ) =

q∗i , γ ∈[

π2, π]+ [γ0, 2π]

−ıθj[x(γ)], γ ∈

[0, π

2

]+ [π, γ0]

.

Since the sought solution is assumed to be bounded throughout the domain Dz, thefunction h(γ) has to satisfy the condition

∫ 2π

0

h(γ)

R (eıγ)dγ = 0 (3.5.4)

which provides boundedness of ω(ζ) at the origin. Therefore ω(ζ) can be written in theform

ω(ζ) =ζR(ζ)

πı

∫ 2π

0

h(γ)

R (eıγ) (eıγ − ζ)dγ.

¿From the last formula we infer

‖ω(ζ)‖Cβ≤ K3 (q∗i , ε1, ‖θ′x(x)‖C) ,

where β = min (α0,12), whence the first part of inequality (3.1.6) (the quantities q∗i will

be defined later on independently on the sought solution ω(τ)).We proceed to proof the unique solvability of the original problem.We map by means of a conformal mapping the domain Dτ of the plane τ = x + ıy

onto the circle |ζ| < 1 so that the points A, B, C correspond to the points 1, ı, −ı, and,since l is not prescribed, the image of the point D is eıγ0 ≡ t0 ≡ t0(l). For the functionω(ζ), satisfying the transformed equation (3.1.12)


ωζ − µ(ζ, ω)ωζ = 0, (3.5.5)

we get then the boundary-value problem

<e (a+ ıb)ω(γ)h1(γ), γ ∈ [0, 2π], (3.5.6)

where µ(ω, ζ) = µ(ω)e−2ı arg dτdζ ,

a+ ıb =

1, γ ∈[

π2, π]+ [γ0, 2π]

,

−ı, γ ∈[

0, π2

]+ [π, γ0]

.

Let us include the solution of problem (3.5.5)-(3.5.6) into the one-parameter familyωλ(ζ) of the solutions of equation (3.5.5) which satisfy the boundary conditions onthe circumference |ζ| = 1

<e (a+ ıb)ωλ(γ) = λh1(γ) ≡ hλ(γ), γ ∈ [0, 2π].

Evidently, in the case λ = 0 there is the unique solution of the starting problemω(z) ≡ 0, which describes the unperturbed plane motion, and the case λ = 1 correspondsto the solution of the problem for the curves AOB, COD prescribed. It is clear that forthe solutions of the problems of overflow of the curves (AOB), (COD) with the inclinationangles θλi = λθi(x) inequalities (3.1.15) hold where N and ε can be taken independent ofλ. Equation (3.5.5) is then uniformly elliptic on the sought solutions , that is

|µ(ωλ, ζ)| ≤ µ0 < 1 (3.5.7)

where µ0 does not depend on λ. Let us seek solutions of problems (3.5.5)-(3.5.6) in theform

ωλ =R0(σ)

π

∫|σ|

eıγh [σ (eıγ)]

R0 (eıγ) (eıγ − σ)dγ ≡ F λ(σ),

where σ = σ(λ, ζ) is the sought homemorphism of the circle |ζ| < 1 on the circle |σ| < 1realized by the solution of the equation

σζ − µ[F λ(σ), ζ

]σζ = 0

and normalized by the conditions σ(0) = 0, σ4 = σ(t0) = t0, R0(σ) = Π4k=1(σ − σk)

1/2,and σ1(1) = σ1, σ2 = σ(ı), σ3 = σ(−1). Besides, the function ωλ = F λ(σ) solves theboundary-value problem only under the additional condition

ah ≡∫ 2π

0

h(γ)eıγ

R0 (eıγ)dγ = 0. (3.5.8)

On the other hand,


ah = q∗1

∫ σ3

σ2

eıγ

R0 (eıγ)dγ +

∫γ∈[σ2,σ3]

h(γ)eıγ

R0 (eıγ)dγ ≡ J(t0)q

∗1 −B1(t0, q

∗2) = 0,

and, moreover, the integrand of J(t0) is nonnegative. Thus, the value of q∗1 can becalculated via the formula

q∗1 =B1(t0, q

∗2)

J(t0)= B(t0, q

∗2).

By virtue of the already proven inequality (3.5.7) the homeomorphism σ = σ(ζ) is a Holderhomeomorphism with the constant α depending only on µ0, therefore ωλ = F λ[σ(ζ)] ∈Cα0 .

The homeomorphism σ = σ(ζ) will be sought in the form

σ = Φ[ζ+·µ ζ]eT f ≡ Φ(Z)eT0f [ζ]−T0f [t0],

where·µ= µ[F (0), 0],

T0f = − 1

π

∫ ∫K

[f(t)

t− ζ+ζf(t)

1− ζt

]dKt,

and Φ(Z) is an analitic function which realizes a conformal mapping of the ellips of the

plane Z with the ratio of semi-axis equal to 1+·µ

1−·µ

onto the circle |σ| < 1, and normed by

the conditions Φ(0) = 0 and Φ[Z(t0)] = t0. To define the function f ∈ Lp, p > 2 we getthe following nonlinear singular equation

f − µ[F λ(Φ), ζ

]S0f =

[µ[F λ(Φ), ζ]−

·µ]ΦZ

Φ[ζ−·ζ]

, (3.5.9)

where S0f = ∂T0f∂ζ

is a linear bounded in Lp operator, and ‖S0|L1 = 1. Equation (3.5.9) isreduced to the equation

f = (I − µS0)−1 g ≡ Aλf. (3.5.10)

Similarly to §4, we establish that for every fixed λ the operator Aλ is defined and com-pletely continuous on the function set f ∈M ∈ Lp :∣∣∣=m

F λ(Φ)∣∣∣ < π

2− ε0, ε0 > 0,

∣∣∣<e F λ(Φ)∣∣∣ < N <∞,

and that it is equicontinuous in λ for every fixed f ∈ M. It follows from inequalities(3.1.15) that the boundary of the set M doesn’t contain a fixed point of the operator


Aλ. Performing a continuous extension of the operator Aλ on the whole of the space,and applying to equation (3.5.10) the Leray-Schauder principle of the fixed point, weconclude that for every λ ∈ [0, 1] there exists at least one solution of equation (3.5.10).Under condition (3.5.8), for λ = 1 we conclude that there exists at least one solutionto the original pronblem. The uniqueness of this solution follows from Lemma 2 of §2,Chapter 1. Note that condition (3.5.8) is a condition on the parameters t0 = t0(l), q

∗2,

and on θi(x) which are the inclination angles of the prescribed arcs.

3.6 The problem of overflow of a heavy incompress-

ible fluid over a spillway

In this problem the spillway profile AOBCD is prescribed. It is requested to define theequation of the free boundary AE and the distribution of the complex velocity of the flowdwdz

= q(z)e−ıθ(z) in the domain Dz which is bounded by the curves AED and AOBCD.

Under the assumption that θ(x) = arctan dydx

has a constant sign along the boundaryAOBC, the theorem of existence of a solution to a similar problem was proved by Jerbay[35]. The boundary conditions of the problem can be written in the form

ψ = 0 on AOBCDy = y(x) or θ = θ(x), .

(3.6.1)

ψ = q∞H∞q2 = 2gy + q2

∞ − 2gH∞on AED.

Let us reduce this problem to a nonlinear boundary-value problem for a quasilinear


equation. We proved unable to show the existence of a solution to this problem andconstructed an approximate numerical solution [17].

Let us take into consideration the plane τ = x + ıψ. The image of the flow domainDz in this plane is the strip 0 ≤ ψ ≤ q∞H∞ (fig.)


.

On the line ψ = 0 the function y = y(x) is given, and q2 = 2gy + q2∞ − 2gH∞ on

the straight line ψ = q∞H∞. In the case of incompressible fluid, the function ω = ln dwdz

satisfies on the plane τ = x+ ıψ the equation

ωτ −eω − 1

eω + 1ωτ = 0 (3.6.2)

which is equivalent to the system

ux = vuψ − uvψ,

vx = uuψ + vvψ, (3.6.3)

u =∂φ(x, y)

∂x, v =

∂φ(x, y)

∂y.

Let us deduce the equation for the function y = y[x(φ, ψ), ψ]. We have

∂y(φ, ψ)

∂φ= yx

∂x(φ, ψ)

∂φ,

yψ =∂y(φ, ψ)

∂ψ= yx

∂x(φ, ψ)

∂ψ,

whence, passing to the inverse functions φ = φ(x, y), ψ = ψ(x, y), we get

yx =v

u, yψ =

1

u. (3.6.4)

Making use of (3.6.4), we can rewrite system (3.6.3) in the following form:

q2 =1 + y2

x

y2ψ

, 2

(yxyψ

)x

=(q2)ψ, q2 = u2 + v2 =

1 + y2x

y2ψ

. (3.6.5)


Excluding q2 from system (3.6.5), we arrive at the following problem for the functiony = y(x, ψ):

y2ψyxx − 2yψyxyψx + (1 + y2

ψ)yψψ = 0, (3.6.6)

y = y(x) on the straight line ψ = 0, (3.6.7)

1 + y2x

y2ψ

= 2gy + q2∞ − 2gH∞ on the line ψ = q∞H∞. (3.6.8)

To construct an approximate solution we assumed that far enough from the spillwayup and down the stream the conditions

y = kiψ (3.6.9)

were fulfilled on the truncations x = xi, i = 1, 2, with ki found from the condition thatBernulli’s integral held on the free boundary.

The derived nonlinear problem (3.6.6)-(3.6.9) was solved in the rectangular 0 ≤ ψ ≤q∞H∞, x1 ≤ x ≤ x2 via the iterational method of fractional steps [38]. At each fractionalstep t = (n+ 1

2)τ the equation

1

2yt =

1 + (ynx)2(

ynψ)2 yψψ − 2

ynxynψynxψ (3.6.10)

is solved, where ynx , ynψ, ynxψ and the initial conditions are taken from the previous integer

step, and under the boundary conditions

y = y(x) for ψ = 0,y = yn(x) for ψ = q∞H∞.

(3.6.11)

At each integer step t = (n+ 1)τ the equation

1

2yt = yxx (3.6.12)

is solved under the initial conditions taken from the previous fractional step, and underthe boundary conditions

y = kiψ at x = xi. (3.6.13)

The corresponding finite-difference equations are solved by the three-point sweepmethod. Thereafter the boundary values of the function yn in (3.6.11) are recalculatedfrom the boundary condition

αyt = ynxyx − (C − 2gyn)ynψyψ. (3.6.14)


In this process either a given number of iterations is performed, or it is interruptedonce the prescribed exactness is achieved:

maxx

∣∣∣yn+1(x) − yn(x)

∣∣∣ ≤ ε.

The calculations performed have exposed the convergence of this process. The numer-ical experiment was realized for the case where

y(x) =1

β2 + x2(3.6.15)

under the values of the parameters x1 = −5m, x2 = 5m, H = 1m, and with various ki.The following forms of the free boundary were obtained:

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