ON BVPS FOR STRONGLY ELLIPTIC SYSTEMS WITH HIGHER ORDER BOUNDARY CONDITIONS · 2007. 2. 23. · Representation (1.3) can also be used for solving boundary value problems with boundary

Georgian Mathematical JournalVolume 14 (2007), Number 1, 145–167

ON BVPS FOR STRONGLY ELLIPTIC SYSTEMS WITHHIGHER ORDER BOUNDARY CONDITIONS

FLAVIA LANZARA

Dedicated to the memory of Professor G. Fichera

Abstract. We consider BVPs for strongly elliptic systems of order 2l withthe boundary conditions of order l + n, n > 0. By representing the solutionby means of a simple layer potential of order n and by imposing the boundaryconditions, we get a singular integral system which is of regular type if andonly if the boundary operator satisfies the Lopatinskĭı condition and whichcan be solved if suitable compatibility conditions are satisfied. An explicitformula for computing the index of the BVP is given.

2000 Mathematics Subject Classification: Primary: 35J55. Secondary:45F15, 35G15.Key words and phrases: Strongly elliptic system, simple layer potential,higher order boundary conditions, Lopatinskĭı condition.

1. Introduction

Let us consider the Dirichlet problem for the Laplace equation in a boundeddomain A ⊂ R2 with a Lyapunov boundary

∆2u = 0 in A, u = f on ∂A. (1.1)

The classical way of solving this problem is to represent the solution by meansof a double layer potential

u(z) =

∫

∂A

ϕ(ζ)∂

∂νζlog |z − ζ| dsζ (1.2)

( ∂∂νζ

denotes differentiation along the inward normal at the point ζ of ∂A; sζis the arc length) where ϕ has to be determined. By imposing the boundarycondition u = f on ∂A we get a Fredholm integral equation

−π ϕ(z) +∫

∂A

ϕ(ζ)∂

∂νζlog |z − ζ| dsζ = f(z), z ∈ ∂A,

which has one and only one solution ϕ in C0(∂A) for any right-hand side f inthe same space. We remark that in this case (1.2) belongs to C0(A).

The double layer potential approach has been extended, in two different ways,in [1] and in [4] to higher order elliptic operators with constant coefficients.

ISSN 1072-947X / $8.00 / c© Heldermann Verlag www.heldermann.de

146 F. LANZARA

A second method of solving (1.1), via integral equations, consists in repre-senting u by a simple layer potential

u(z) =

∫

∂A

ϕ(ζ) log |z − ζ| dsζ . (1.3)

When we impose the boundary condition u = f on ∂A we get the followingintegral equation of first kind on ∂A∫

∂A

ϕ(ζ) log |z − ζ| dsζ = f(z) . (1.4)

If we suppose that f ∈ C1+λ(∂A), differentiating (1.4) with respect to the arclength leads to the integral equation∫

∂A

ϕ(ζ)∂ log |z − ζ|

∂szdsζ =

∂f(z)

∂sz.

Here the integral has to be understood as a singular integral, due to the strongsingularity of the kernel. As it was shown by Muskhelishvili, this equation canbe regularized to a Fredholm equation. The solution of the Dirichlet problemobtained by this method belongs to C1+λ(A).

In [6] the extension of this method to higher order elliptic operators has beenconsidered.

Representation (1.3) can also be used for solving boundary value problemswith boundary conditions of first order

b1(z)ux + b2(z)uy + c(z)u = f on ∂A (1.5)

satisfying the condition b21(z)+b22(z) 6= 0 (equivalent to the so-called Lopatinskĭı

condition). Of course, in this case f has to satisfy a finite number of compati-bility conditions. If we impose the boundary condition (1.5) to the simple layerpotential (1.3), we get a singular integral equation which is of regular type andwhich can be solved for any f satisfying the above-mentioned compatibilityconditions.

Suppose we look for a more regular solution u ∈ C1+n+λ(A), n > 0, of aboundary value problem with higher order boundary conditions, e.g.,

∆2u = 0 in A,2∑

j=0

dj(z)∂2u

∂x2−j∂yj+ b1(z)ux + b2(z)uy + c(z)u = f on ∂A.

(1.6)

Problem (1.6) contains, as a very particular case, the Laplace equation with theboundary condition

∂2u

∂ν2ζ+ p

∂u

∂νζ= f on ∂A,

p(z) ∈ Cλ(∂A), which was solved by means of a different kind of potentials in[13]. Also, the Ventssel boundary conditions fall into (1.6). This problem was

ON BVPS FOR STRONGLY ELLIPTIC SYSTEMS 147

originally stated in [22] and considered by many authors in the recent years(see, e.g., [5], [3] and the references therein).

In order to solve problem (1.6) we use the concept of simple layer potentialof order n. These potentials were introduced in [10]. There the authors statedthat a solution of class Cn+l+λ(A) of a general elliptic system of order 2l in theplane could be represented by means of such potentials. This result was laterproved in [11].

In the present paper we study the boundary value problem

Eu = 0 in A; Bu = f on ∂A, (1.7)

where E is a q×q matrix differential operator of order 2l with constant complexcoefficients and B is a matrix differential operator of order l+n, n > 0, on ∂A. Eis supposed to be strongly elliptic. The function f belongs to the space Cλ(∂A)and we look for a solution u in the space Cn+l+λ(A). In order to investigateproblem (1.7) we represent u by a simple layer potential of order n

uν(z) =

q∑µ=1

l−1∑j=0

∫

∂A

ϕµj(ζ)∂l−1

∂ξl−1−j∂ηjF (n)νµ (z, ζ)dsζ , ν = 1, . . . , q, (1.8)

where {F (n)νµ (z, ζ)} is defined in Section 2.The paper is organized as follows. In Section 2 we recall the definition of

a fundamental solution of order n, the definition of a simple layer potential oforder n in the sense of Fichera and Ricci and the Representation Theorem forsolutions of the strongly elliptic system Eu = 0 in A. By imposing, on thefunction u given by (1.8), the boundary conditions we get a system of singularintegral equations. In Section 3 we prove that this system is of regular type ifand only if the operator B satisfies the Lopatinskĭı condition with respect to theoperator E. Our proof rests on some results in the theory of matrix polynomialsobtained in [23] and [18]. The regularity of the singular integral system impliesits solvability provided that suitable compatibility conditions are satisfied. Thismeans that we obtain the solutions of (1.7) by means of a simple layer potentialsof order n. In Section 4 an explicit formula for computing the index of the BVP(1.7) is given.

2. A Simple Layer Potential of order n

Let A be a simply connected open set of the plane of the complex variablez = x+ iy such that ∂A has a uniformly Hölder continuous normal field of someexponent λ, (0 < λ 6 1), (∂A ∈ C1+λ).

Let l and q be positive integers and m = lq. In the following we consider thespaces:

• Cλ(∂A) formed by all complex functions satisfying, on ∂A, a uniformHölder condition with some Hölder exponent λ : 0 < λ 6 1;

• C l+λ(A) formed by all complex functions which are continuously differ-entiable up to the order l with respect to the real variables x and y, and

148 F. LANZARA

each partial derivative of order l is uniformly Hölder continuous in Awith some exponent λ;

• L2(∂A) formed by all measurable complex functions such that |u|2 isintegrable over ∂A;

• [Cλ(∂A)]q, [C l+λ(A)]q and [L2(∂A)]q formed by all vector functions Φ =(ϕ1, . . . , ϕq) such that ϕi ∈ Cλ(∂A), ϕi ∈ C l+λ(A) and ϕi ∈ L2(∂A)respectively;

• H = [L2(∂A)]m × Cσn , σn = q(2ln + l(l+1)

2

), formed by the pairs (Φ; c)

such that Φ ∈ [L2(∂A)]m and c ∈ Cσn . We set < (Φ; c), (Ψ; d) >=(Φ, Ψ) +

σn∑i=1

cidi, where (Φ, Ψ) =m∑

i=1

∫∂A

ϕiψids.

• B = [Cλ(∂A)]m×Cσn formed by the pairs (Φ; c) such that Φ ∈ [Cλ(∂A)]mand c ∈ Cσn .

We consider systems of linear differential equations E u = 0 in A where u isa q-vector function and E = {Eµν} is a q × q matrix such that the elementsEµν are linear differential operators

Eµν = Eµν(

∂

∂x,

∂

∂y

)=

2l∑

k=0

aµνk∂2l

∂x2l−k∂yk, µ, ν = 1, . . . , q,

with constant complex coefficients ak = {aµνk }µ,ν=1,...,q.We denote by

{Eµν} ={

Eµν

(∂

∂x,

∂

∂y

)}, µ, ν = 1, . . . , q,

the matrix obtained by taking the matrix of the co-factor of the matrix E and

by transposing it. Let L(w) =2l∑

k=0

akw2l−k be the q × q matrix polynomial of

order 2l associated to the differential operator E and let Lµν(w) and Lµν(w)be the characteristic polynomials associated to the differential operators Eµν

and Eµν , respectively , i.e., Lµν(w) = Eµν(w, 1) and Lµν(w) = Eµν(w, 1). If

L̃(w) = {Lµν(w)}µ,ν=1,...,q we have L(w) L̃(w) = L̃(w) L(w) = det L(w)I.Hypothesis 1: The operator E is elliptic in the sense of Petrovskii, i.e.,

det L(w) 6= 0, ∀w ∈ R and det a0 6= 0.Hypothesis 2: The polynomial det L(w) of degree 2m (m = lq) has m zeroes

with the positive imaginary part and m zeroes with the negative imaginary part.Let us denote by Γ a rectifiable closed Jordan curve in the complex plane

w, lying in the half plane =w < 0, such that the bounded domain which hasΓ as contour contains all the zeroes of det L(w) which belong to the half plane=w < 0. Let Γ∗ be the symmetric of Γ with respect to the real axis. Supposethat the bounded domain which has Γ∗ as contour contains all the zeroes of


det L(w) which belong to the half plane =w > 0. Consider the function

Pα(z, ζ) = −14π2α!

∫

+Γ

[(x− ξ)w + (y − η)]α log[(x− ξ)w + (y − η)]det L(w)

dw

+1

4π2α!

∫

+Γ∗

[(x− ξ)w + (y − η)]α log[(x− ξ)w + (y − η)]det L(w)

dw,

α denoting a non-negative integer. The determination of log[xw + y] on Γ andΓ∗ is chosen as specified in [10].

Let n be a non-negative integer. Consider the following q × q matrix:F (n)(z, ζ) = {F (n)µν (z, ζ)}µ,ν=1,...,q ,

where

F (n)µν (z, ζ) = Eµν(∂

∂x,

∂

∂y)P2m+n−2(z, ζ), µ, ν = 1, . . . , q.

We call F (n)(z, ζ) the fundamental solution matrix of order n for the operator E.Since Eµν is a differential operator of order less than or equal to 2m− 2l, by

using the same technique as employed in [6, pp. 65–66], essentially based on thePlemelij formula, we get

Theorem 2.1. If ω ∈ Cλ(∂A), for any z0 ∈ ∂A the following limit relationholds

limz→z0

∫

∂A

ω(ζ)∂2l+n−1

∂x2l+n−k−1∂ykF (n)νσ (z, ζ) dsζ

=iω(z0)

4π

{ ∫

+Γ

Lνσ(w)w2l+n−1−k

det L(w)(ẋ0w + ẏ0)dw +

∫

+Γ∗


det L(w)(ẋ0w + ẏ0)dw

}

− 14π2

{ ∫

∂A

ω(ζ)dsζ

[ ∫

+Γ


det L(w)[(x0 − ξ)w + (y0 − η)] dw

−∫

+Γ∗


det L(w)[(x0 − ξ)w + (y0 − η)] dw]}

, k = 0, . . . , 2l + n− 1,

as z tends to z0 in the interior of A (the dot denotes differentiation with respectto the arc length on ∂A, oriented in the counterclockwise sense). The integralover ∂A must be understood as a singular Cauchy integral.

If Φ = {ϕσj}, (σ = 1, . . . , q; j = 0, . . . , l − 1) belongs to [Cλ(∂A)]m, definev0[Φ] = (v01[Φ], . . . , v

0q [Φ]):

v0ν [Φ](z) =

q∑µ=1

l−1∑j=0

∫

∂A

ϕµj(ζ)∂l−1

∂ξl−1−j∂ηjF (n)νµ (z, ζ)dsζ , ν = 1, . . . , q. (2.1)

150 F. LANZARA

The function v0[Φ] is called a simple layer potential of order n and

v0[Φ] ∈ V (A) = {u ∈ [C2l(A)]q ∩ [C l+n+λ(A)]q : Eu = 0 in A}

(see [10] and, for a complete proof, [11, Theorem 2.5]).

The matrix differential operator E is said to be a strongly elliptic operator if,for any real w and for every non-zero complex vector η = (η1, . . . , ηq), we have

<(

1,q∑i,j

Lij(w)ηj η̄i

)> 0; <

(1,q∑i,j

aij0 ηj η̄i

)> 0.

We shall assume E to be strongly elliptic. Then hypotheses 1), 2) are satisfied(see [2, p. 43], [8, p. 101], [14, p. 275], [16, p. 669],[23, p. 425]).

For any integer n > 0, consider the followingProblem (Pn): find u ∈ [Cn+l(A)]q such that

Eu = 0 in A,

∂

∂s

∂n+l−1

∂xl−1−h∂yh+nu = 0 on ∂A, h = 0, . . . , l − 1.

Theorem 2.2. For a fixed n > 0, there exist

σn = q

[2ln +

l(l + 1)

2

]

linearly independent solutions of problem (Pn).

Before proving Theorem 2.2, we start with a preliminary result:

Proposition 2.1. Let ω = (ω1, . . . , ωq) be a polynomial solution of the systemEω = 0 in A. Then there exist polynomials ω̃ = (ω̃1, . . . , ω̃q) such that

Eω̃ = 0 ;∂ω̃

∂y= ω. (2.2)

Proof. If ω is a polynomial solution of the system Eω = 0, then

χ(x) := E

( y∫

y0

ω(x, η)dη

)(2.3)

is a polynomial in x. Indeed,[E

( y∫

y0

ω(x, η)dη

)]

µ

=

q∑ν=1

{aµν0

y∫

y0

∂2l

∂x2lων(x, η)dη +

2l∑

h=1

aµνh∂2l−1

∂x2l−h∂yh−1ων(x, y)

}


=

q∑ν=1

2l∑

h=1

aµνh

{−

y∫

y0

∂2l

∂x2l−h∂ηhων(x, η)dη +

∂2l−1


}

=

q∑ν=1

2l∑

h=1

aµνh∂2l−1


∣∣∣y=y0

, µ = 1, . . . , q.

On the other hand, the condition det a0 6= 0 ensures the existence of a poly-nomial solution v(x) of the system Ev = χ, i.e., a0

∂2l

∂x2lv(x) = χ(x). We obtain

that the polynomial

ω̃(x, y) =

y∫

y0

ω(x, η)dη − v(x)

satisfies (2.2). ¤

Proof of Theorem 2.2. First we prove, by induction, that (Pn) has only polyno-mial solutions. It is known that (P0) has as solutions q-valued polynomials of

degree 6 l − 1: u(x, y) =0,l−1∑j+k

cj,kxj yk, cj,k ∈ Cq. Since the number of linearly

independent monomials of the form cj,kxjyk, 0 6 j + k 6 l − 1, is l(l + 1)/2,

it follows that σ0 = ql(l+1)

2. Let us suppose that (Pn) has only polynomial so-

lutions. If u is a solution of (Pn+1), then ω :=∂u∂y

is a polynomial solution of

(Pn). Hence u has the representation

u(x, y) =

y∫

y0

ω(x, η)dη + ϕ(x).

In the proof of Proposition 2.1 we showed that χ in (2.3) is a polynomial in xand, since Eu = Eϕ + χ = 0, i.e., a0ϕ

(2l)(x) + χ(x) = 0, we obtain that ϕ, too,is a polynomial.

Let us suppose that (Pn) has σn linearly independent solutions {ω(i)}. If uis a solution of (Pn+1),

∂u∂y

is solution of (Pn) and∂u∂y

=σn∑i=1

ciω(i). It implies

(Proposition 2.1)

u =σn∑i=1

ciω̃(i) + Φ(x), Φ = (ϕ1, . . . , ϕq),

where Φ is a q-valued polynomial because u is a q-valued polynomial. Moreover,since

Eu = EΦ = a0∂2l

∂x2lΦ(x) = 0 in A

and det a0 6= 0, ϕi(x) is a polynomial such that deg ϕi 6 2l − 1.

152 F. LANZARA

Let us denote by {p(h)}h=1,...,2lq the polynomials(1, 0, . . . , 0)︸︷︷︸

q

; . . . ; (0, . . . , 0, 1)︸︷︷︸q

; (x, 0, . . . , 0)︸︷︷︸q

; . . . (0, . . . , 0, x2l−1)︸︷︷︸q

.

Then Φ(x) =2lq∑h=1

γhp(h)(x), γh ∈ C. We claim that {ω̃(h)}h=1,...,σn and

{p(h)}h=1,...,2lq are linearly independent. Ifσn∑i=1

ciω̃(i)(x, y) +

2lq∑

h=1

γhp(h)(x) = 0

we haveσn∑i=1

ci∂

∂yω̃(i)(x, y) =

σn∑i=1

ciω(i)(x, y) = 0

which implies c1 = · · · = cσn = 0. Then γ1 = · · · = γ2lq = 0.We deduce that σn+1 = σn + 2lq. ¤Theorem 2.3 (Representation Theorem). Suppose that the operator E is

strongly elliptic and ∂A is of class C1+λ with λ > 1/2. u belongs to V (A) if andonly if there exists (Φ; c) ∈ B such that

u = v0[Φ] + p[c], (2.4)

where v0[Φ] is defined in the right-hand side of (2.1) and

p[c] =σn∑i=1

ciω(i), (2.5)

{ω(j)}j=1,...,σn being linearly independent polynomial solutions of (Pn).A detailed proof of Theorem 2.3 is contained in [11]. Here we recall some

results which will be used in the next sections.Consider a function u ∈ V (A) and set

gh(z) =∂

∂sz

∂l−1

∂xl−1−h∂yh∂n

∂ynu(z), 0 6 h 6 l − 1.

We want to determine Φ in such a way that

∂

∂sz

∂l−1

∂xl−1−h∂yh∂n

∂ynv0[Φ](z) = gh(z), h = 0, . . . , l − 1, z ∈ ∂A. (2.6)

Using Theorem 2.1 the system (2.6) may be rewritten in the canonical form as

(−1)l−14πgνh(z) =q∑

σ=1

l−1∑j=0

{Aνh,σjϕσj(z) + Bνh,σj

πi

∫

+∂A

ϕσj(ζ)

ζ − z dζ

+

∫

+∂A

ϕσj(ζ)Mνh,σj(z, ζ)dζ}

, ν = 1, . . . , q; h = 0, . . . , l − 1, z ∈ ∂A, (2.7)


where gh = {gνh}ν=1,...,q;

Aνh,σj = i[∫

+Γ

Lνσ(w)

det L(w)w2l−(h+j)−2 dw +

∫

+Γ∗

Lνσ(w)

det L(w)w2l−(h+j)−2 dw

];

Bνh,σj = i[∫

+Γ

Lνσ(w)

det L(w)w2l−(h+j)−2 dw −

∫

+Γ∗

Lνσ(w)


];

Mνh,σj(z, ζ) = 1π

[ ∫

+Γ

Lνσ(w)H(w, z, ζ)


−∫

+Γ∗

Lνσ(w)H(w, z, ζ)


]= O (|z − ζ|λ−1) , 0 < λ 6 1.

Here H(w, z, ζ) = K(w, z, ζ)− 1, where

K(w, z, ζ) =

ζ̇(ẋw + ẏ)

x− ξz − ζ w +

y − ηz − ζ

, z 6= ζ,

1, z = ζ

defined for w ∈ Γ ∪ Γ∗ and (z, ζ) ∈ ∂A× ∂A.System (2.7) is a system of singular integral equations of the kind

g(z) = AΦ(z) + B(SΦ)(z) + (MΦ)(z), z ∈ ∂A, (2.8)where g = {gνh}; A, B and M are the matrices of order lq: {Aνh,σj}, {Bνh,σj}and {Mνh,σj} (h = 0, . . . , l − 1; ν = 1, . . . , q), (j = 0, . . . , l − 1; σ = 1, . . . , q);SΦ is the singular integral operator

(SΦ)(z) =1

πi

∫

+∂A

Φ(ζ)

ζ − z dζ

and MΦ is the Fredholm integral operator

(MΦ)(z) =

∫

+∂A

M(z, ζ) Φ(ζ) dζ.

System (2.8) is of regular type, according to Muskhelishvili theory (see [11,p.263]). It follows that, for system (2.8), the same Fredholm theorems of regularintegral systems hold.

Let us consider the associated homogeneous system

0 = AΦ(z) + B(SΦ)(z) + (MΦ)(z), z ∈ ∂A, (2.9)

154 F. LANZARA

and the adjoint homogeneous system, which we can write as follows:

0 =

q∑ν=1

l−1∑

h=0

∫

∂A

ψνh(ζ)∂

∂sζ

∂l−1

∂ξl−1−h∂ηh∂l−1

∂xl−1−j∂yjF (0)νσ (ζ, z)dsζ , (2.10)

j = 0, . . . , l − 1; σ = 1, . . . , q, z ∈ ∂A.System (2.10) has ql(l + 1)/2 linearly independent solutions. Denoting by

χ the index of system (2.8), i.e., the difference between the dimensions ofthe spaces of solutions of (2.9) and (2.10), respectively, we have χ = 0 ([15,p.419],[17, p.14],[9, p.5]). Moreover, (2.8) has solutions if and only if the datumg is orthogonal to the eigensolutions of (2.10). These compatibility conditionsare always satisfied as it is shown in [11]. If Φ is any solution of (2.8), the simplelayer potential (2.1) gives a solution of Eu = 0 in A determined up an arbitrarysolution of (Pn).

3. Boundary Value Problem

Let us denote by B = {Bµν}, µ = 1, . . . , m, ν = 1, . . . , q, a m × q matrixboundary operator. It is taken in the form

B = B0 + B1, B0 = {Bµν0 }, B1 = {Bµν1 },where

Bµν0 =l+n∑

h=0

bµνh (z)∂l+n

∂xl+n−h∂yh; Bµν1 =

l+n−1∑s=0

l+n−1−s∑i=0

bµνis (z)∂l+n−1−s

∂xl+n−1−s−i∂yi.

The functions bµνh (z) and bµνis (z) belong to C

λ(∂A).For a strongly elliptic operator E, we consider the following

BVP: Given f ∈ [Cλ(∂A)]m, find u ∈ [Cn+l+λ(A)]q ∩ [C2l(A)]q:Eu = 0 in A; Bu = f on ∂A.

Denote by B0(z; w) = {Bµν0 (z; w)} the following lq × q matrix polynomial:

B0(z; w) =l+n∑

h=0

bh(z)wl+n−h, bh(z) = {bµνh (z)}µ=1,...,lq;ν=1,...,q.

Consider the simple layer potential of order n: v0[Φ] defined in (2.1).From Theorem 2.1 we obtain, for any z0 ∈ ∂A,

limz→z0

(B0v0[Φ])µ(z)

= (−1)l−1q∑

σ=1

l−1∑j=0

{iϕσj(z0)

4π

[ ∫

+Γ

q∑ν=1

Bµν0 (z0; w)Lνσ(w)wl−1−j

(ẋ0w + ẏ0)

dw

det L(w)

+

∫

+Γ∗

q∑ν=1


(ẋ0w + ẏ0)

dw

det L(w)

]


− 14π2

∫

∂A

ϕσj(ζ)dsζ

[ ∫

+Γ

q∑ν=1


[(x0 − ξ)w + (y0 − η)]dw

det L(w)

−∫

+Γ∗

q∑ν=1


[(x0 − ξ)w + (y0 − η)]dw

det L(w)

]}.

If H(w, z, ζ) is the function defined in Section 2, w ∈ Γ∪Γ∗; (z, ζ) ∈ ∂A×∂A,since

dsζ(x− ξ)w + (y − η) =

ζ̇dζ

(x− ξ)w + (y − η)=

1

ẋw + ẏ

dζ

z − ζ +1

ẋw + ẏ

H(w, z, ζ)

z − ζ ζ̇ dsζ

we have, for z ∈ ∂A,

(B0v0[Φ])µ(z) =

(−1)l−14π

q∑σ=1

l−1∑j=0

{αµ,σj(z)ϕσj(z)

+1

iπβµ,σj(z)

∫

+∂A

ϕσj(ζ)

ζ − z dζ +1

π

∫

∂A

γµ,σj(z, ζ)ϕσj(ζ)dsζ

}, µ = 1, . . . , m,

where

αµ,σj(z) = i

[ ∫

+Γ

q∑ν=1

Bµν0 (z; w)Lνσ(w)wl−1−j

(ẋw + ẏ)

dw

det L(w)

+

∫

+Γ∗

q∑ν=1


(ẋw + ẏ)

dw

det L(w)

];

βµ,σj(z) = i

[ ∫

+Γ

q∑ν=1


(ẋw + ẏ)

dw

det L(w)

−∫

+Γ∗

q∑ν=1


(ẋw + ẏ)

dw

det L(w)

];

γµ,σj(z, ζ) =

∫

+Γ

q∑ν=1


det L(w)(ẋw + ẏ)ζ̇H(w, z, ζ)

ζ − z dw

−∫

+Γ∗

q∑ν=1


det L(w)(ẋw + ẏ)ζ̇H(w, z, ζ)

ζ − z dw.

156 F. LANZARA

For any w ∈ Γ ∪ Γ∗, H(w, z, ζ)[det L(w)(ẋw + ẏ)]−1 is uniformly Höldercontinuous with respect to (z, ζ) ∈ ∂A× ∂A. Moreover we have, for z ∈ ∂A,

(B1v0[Φ])µ(z) = (−1)l−1

q∑σ=1

l−1∑j=0

∫

∂A

ϕσj(ζ)

×q∑

ν=1

Bµν1

(z;

∂

∂x,

∂

∂y

)∂l−1

∂xl−1−j∂yjF (n)νσ (z, ζ)dsζ , µ = 1, . . . ,m.

Define the matrices having m rows: µ = 1, . . . , m and m columns: σ =1, . . . , q; j = 0, . . . , l − 1

A(z) = {αµ,σj(z)}; B(z) = {βµ,σj(z)}; Q(z, ζ) = {Qµ,σj(z, ζ)},where

Qµ,σj(z, ζ) =1

πγµ,σj(z, ζ) + 4π

∂l−1

∂xl−1−j∂yj

q∑ν=1

Bµν1

(z;

∂

∂x,

∂

∂y

)F (n)νσ (z, ζ).

Qµ,σj(z, ζ) are weakly singular kernels.If Φ = {ϕσj} ∈ [Cλ(∂A)]m, we represent

Bv0[Φ](z) =(−1)l−1

4π

{A(z)Φ(z) + 1

iπB(z)

∫

+∂A

Φ(ζ)

ζ − z dζ

+

∫

∂A

Q(z, ζ)Φ(ζ)dsζ}

, z ∈ ∂A.

Introduce the operator S : [L2(∂A)]m → [L2(∂A)]m

SΦ(z) = A(z)Φ(z) + 1iπB(z)

∫

+∂A

Φ(ζ)

ζ − zdζ +∫

∂A

Q(z, ζ)Φ(ζ)dsζ .

S is a singular integral operator.Let u be a solution of BVP. For the Representation Theorem 2.3 there exists

(Φ; c) ∈ H such that u = v0[Φ] + p[c] where v0[Φ] is a simple layer potential oforder n defined in the right-hand side of (2.1) and p[c] is given in (2.5).

Let L be the operator defined on Cσn as follows:Lc = (−1)l−14πBp[c].

L : Cσn → [L2(∂A)]m is a continuous operator.By imposing the boundary conditions to the function u in (2.4), we obtain

T (Φ; c) = (−1)l−14πf, (3.1)where T (Φ; c) = SΦ + Lc. T : H → [L2(∂A)]m is a continuous operator.

Conversely, let (Φ; c) ∈ H be a solution of (3.1). Since Lc ∈ [Cλ(∂A)]m, wehave SΦ ∈ [Cλ(∂A)]m and this implies Φ ∈ [Cλ(∂A)]m (see [7, p.264, TheoremXXII]).

We state this result in the following


Theorem 3.1. If u is solution of BVP, then u can be represented by meansof (2.4) where (Φ; c) ∈ H is a solution of the singular integral system (3.1).

Conversely, if (Φ; c) ∈ H is a solution of (3.1), the function u defined in (2.4)is a solution of the BVP.

Now we study the solvability of system (3.1) that is equivalent, by Theorem3.1, to the solvability of the BVP.

According to Muskhelishvili theory, the singular integral operator S is ofregular type if, ∀z ∈ ∂A, det{A(z)± B(z)} 6= 0.

These conditions can be written in the equivalent way as

det

{ ∫

+Γ

B0(z; w)L−1(w)[I, . . . , wl−1 I]

dw

ẋw + ẏ

}6= 0, z ∈ ∂A;

det

{ ∫

+Γ∗

B0(z; w)L−1(w)[I, . . . , wl−1 I]

dw

ẋw + ẏ

}6= 0, z ∈ ∂A.

Let us factor the polynomial det L(w) as L+(w)L−(w) where L+(w) [L−(w)]is a polynomial of degree m which has as zeroes all the zeroes of det L(w) suchthat =w > 0 [=w < 0].

We shall say that the boundary operator B satisfies the Lopatinskĭı condi-tion (or the complementing boundary condition) with respect to the differentialoperator E if, for every z ∈ ∂A, the rows of the m × q matrix polynomialB0(z; w)L̃(w) are linearly independent mod L

−(w) and mod L+(w) ([2, p.42],[23,p.422]).

This condition means that if, for z ∈ ∂A, there exists d ∈ Cm such thatdB0(z; w)L̃(w) = L

−(w)M(z; w)

for some 1 × q matrix polynomial M(z; w), then d = 0. Analogously, if wereplace L−(w) by L+(w).

Theorem 3.2. Let E be strongly elliptic. The singular integral operator S isof regular type if and only if the boundary operator B satisfies the Lopatinskĭıcondition with respect to the differential operator E.

In order to prove Theorem 3.2 we shall use the method introduced in [23,Ch.10,pp.416-427] and in [18], where the authors reformulate the Lopatinskĭıcondition in various equivalent forms. The main tool is the spectral theory ofmatrix polynomials. We start with some preliminaries and definitions.

In the following we denote by X a q × lq matrix; by T an lq × lq matrix; byY an lq × q matrix and by I the identity matrix of order lq. By hypothesis,the contours Γ and Γ∗ do not intersect the roots of det L(w) = 0 and the zeroesof det L(w) such that =w < 0 [=w > 0] lie inside Γ [Γ∗]. In the following wealways refer to Γ. In the same way it is possible to consider Γ∗.

Denote by σ(T ) the spectrum of T , i.e., σ(T ) = {w ∈ C : det(Iw − T ) = 0}.Definition 3.1. We say that (X, T, Y ) is a Γ-spectral triple for L(w) if

158 F. LANZARA

i) σ(T ) lies inside Γ;ii) L−1(w) − X(Iw − T )−1Y has an analytic continuation inside Γ as a

matrix function of w;iii) the 2lq × lq matrix

X...

X T 2l−1

is injective;iv) the lq×2lq matrix [Y, . . . , T 2l−1Y ] is surjective or, equivalently, has rank

lq.

The fact that the matrix (Iw − T )−1, as a function of w, is analytic, exceptat the points of σ(T ), enables us to obtain some important results by the useof contour integrals in the complex w-plane.

Let U(T ) be the class of complex-valued analytic functions f such that itsdomain is an open set in the complex plane containing σ(T ).

If D is any bounded (regular) domain such that σ(T ) ⊂ D; D ⊂ dom(f), wedefine the operator matrix f(T ) corresponding to f as

f(T ) =1

2πi

∫

+∂D

f(w)(Iw − T )−1dw.

The integral has a value independent of the particular choice of D.Moreover, for any f, g ∈ U(T ),

f(T )g(T ) = g(T )f(T ) =1

2πi

∫

+∂D

f(w)g(w)(Iw − T )−1dw,

where D is any bounded (regular) domain such that σ(T ) ⊂ D; D ⊂ dom(f)∩dom(g) (see [20, p. 289]).

In particular, since σ(T ) lies inside Γ, we have

T j(aT + bI)−1 = (aT + bI)−1T j =1

2πi

∫

+Γ

wj

aw + b(Iw − T )−1dw, (3.2)

j = 0, 1, . . . ; (a, b) ∈ R2 − {(0, 0)}.Proposition 3.1. Property ii) in Definition 3.1 can be replaced by

ii′)1

2πi

∫

+Γ

wj

aw + bL−1(w)dw = XT j(aT + bI)−1Y,

j = 0, 1, . . . ; (a, b) ∈ R2 − {(0, 0)}.Proof. In view of (3.2), condition ii′) is equivalent to

1

2πi

∫

+Γ

wj

aw + b

[L−1(w)−X(Iw − T )−1Y ] dw = 0, (3.3)


j = 0, 1, . . . ; (a, b) ∈ R2 − {(0, 0)}.If ii) holds true, (3.3) immediately follows. Conversely, from (3.3) we deduce

that1

2πi

∫

+Γ

wj[L−1(w)−X(Iw − T )−1Y ] dw = 0, j = 0, 1, . . . .

Since σ(T ) lies inside Γ, this is equivalent to ii). ¤In [23, p. 27] and [18, p. 3115], the following theorem has been proved:

Theorem 3.3. If Γ is a closed contour not intersecting the roots of det L(w),then there exists a Γ-spectral triple for L(w).

If (X, T, Y ) is a Γ-spectral triple for L(w), then σ(T ) coincides with the rootsof det L(w) inside Γ, i.e., with the roots of L−(w) (see [18, p.3116]).

By virtue of Proposition 3.1 the lq × lq matrix∫

+Γ

I...

wl−1 I

L−1(w)[I, . . . , wl−1 I] dw (3.4)

can be written as

X...

X T l−1

[Y, . . . , T l−1Y ]. (3.5)

If E is strongly elliptic, (3.4) is invertible [11, p. 263] which implies that bothlq × lq matrices in (3.5) are nonsingular.

Theorem 3.4. Let us suppose that the lq× lq matrix (3.4) is invertible. Thefollowing statements are equivalent:

a) If (X,T, Y ) is a Γ-spectral triple for L(w) then the lq × lq matrix

∆B(z) =l+n∑j=0

bj(z)XTl+n−j, z ∈ ∂A,

is invertible, i.e., det ∆B(z) 6= 0, z ∈ ∂A.b) The lq × lq matrix

1

2πi

∫

+Γ

B0(z; w)L−1(w)[I, . . . , wl−1 I]

dw

ẋw + ẏ, z ∈ ∂A, (3.6)

is invertible.c) The rows of the lq×q matrix B0(z; w)L̃(w) are linearly independent modulo

L−(w).

Proof. a) ⇔ b) Let (X, T, Y ) be a Γ-spectral triple for L(w). By formula ii′)matrix (3.6) is equal to(

l+n∑j=0

bj(z)XTl+n−j

)(ẋT + ẏI)−1[Y, . . . , T l−1 Y ] (3.7)

160 F. LANZARA

and we know that [Y, . . . , T l−1 Y ] is invertible because (3.4) is invertible. Hence,since σ(T ) lies inside Γ, matrix (3.7) has a nonzero determinant if and only if

∆B(z) =l+n∑j=0

bj(z)XTl+n−j has a nonzero determinant.

b) ⇒ c) If d ∈ Cm is such thatdB0(z; w)L̃(w) = L

−(w)M(z; w) (3.8)

for some 1 × q matrix polynomial M(z; w), then dividing (3.8) by det L(w)(ẋw + ẏ) gives

dB0(z; w)L−1(w)(ẋw + ẏ)−1 =

M(z; w)

L+(w)(ẋw + ẏ)

and the right-hand side is holomorphic inside Γ. Whence

d

∫

+Γ

B0(z; w)L−1(w)

wj

ẋw + ẏdw = 0

and then

d

∫

+Γ

B0(z; w)L−1(w)[I, . . . , wl−1 I]

dw

ẋw + ẏ= 0.

Thus b) implies d = 0.c) ⇒ b) Conversely, let d ∈ Cm such that

d

∫

+Γ

B0(z; w)L−1(w)[I, . . . , wl−1 I]

dw

ẋw + ẏ= 0.

In view of (3.7), since the lq× lq matrix [Y, . . . , T l−1 Y ] is invertible we see thatd∆B(z)(ẋT + ẏI)

−1 = 0.

Multiplying both sides of this equation by T kY for k = 0, 1, . . . and making useof the formula ii′), we obtain

d

∫

+Γ

B0(z; w)L−1(w)

wkdw

ẋw + ẏ= 0, k = 0, 1, . . . .

Hence dB0(z; w)L−1(w) has an analytic continuation inside Γ as a matrix func-

tion of w. This means that dB0(z; w)L̃(w) vanishes at the roots of L−(w) = 0.

Hence (3.8) holds for some 1×q matrix polynomial M(z; w) and then c) impliesd = 0. ¤

If we repeat the same arguments for Γ∗, we deduce Theorem 3.2 from Theo-rem 3.4.

Set

Hµσ(z; w) =

q∑ν=1

Bµν0 (z; w)Lνσ(w), µ = 1, . . . ,m; σ = 1, . . . , q.

Let us regard Hµσ(z; w) as polynomials in the indeterminate w.


For z ∈ ∂A, we reduce the polynomials Hµσ(z; w) mod L−(w) (or modL+(w)):

Hµσ(z; w) ≡m−1∑

β=0

qσβµ (z)wm−1−β mod L−(w) (mod L+(w)).

Then consider the matrix Q(z) = {qσβµ (z)} having m rows: µ = 1, . . . ,m and mqcolumns: β = 0, . . . , m−1; σ = 1, . . . , q. A method for checking the Lopatinskĭıcondition is

Proposition 3.2. The boundary operator B satisfies the Lopatinskĭı condi-tion with respect to the differential operator E if and only if, for every z ∈ ∂A,the rank of Q(z) is m.

Proof. Since, for z ∈ ∂A and σ = 1, . . . , q,m∑

µ=1

dµHµσ(z; w)

≡m−1∑

β=0

(m∑

µ=1

dµqσβµ (z)

)wm−1−β mod L−(w) (mod L+(w)), (3.9)

we havem∑

µ=1

dµHµσ(z; w) ≡ 0 if and only ifm∑

µ=1

dµqσβµ (z) = 0, σ = 1, . . . , q; β = 0, . . . ,m− 1.

The above-considered system has nontrivial solutions if and only if the rank ofQ(z) is less than m. Then, for (3.9), the Lopatinskĭı condition holds true if andonly if the rank of Q(z) is m. ¤

Let T ∗ : [L2(∂A)]m → H be the adjoint operator of T . Then, if (Φ; c) ∈ Hand Ψ ∈ [L2(∂A)]m, we have

〈(Φ; c), T ∗Ψ〉 = (T (Φ; c), Ψ) = (SΦ, Ψ) + (Lc, Ψ)

= (Φ,S∗Ψ) + (−1)l−14πσn∑i=1

ci(Bω(i), Ψ),

S∗ : [L2(∂A)]m → [L2(∂A)]m denoting the adjoint operator of S. ThenT ∗Ψ = (S∗Ψ; (−1)l−14π(Bω(1), Ψ), . . . , (−1)l−14π(Bω(σn), Ψ)) .

Consider the following homogeneous transposed system of (3.1):

T ∗Ψ = 0. (3.10)Theorem 3.5. Suppose that the boundary operator B satisfies the Lopatinskĭı

condition with respect to the differential operator E. Then the BVP admits asolution if and only if

(f, Ψ) = 0, ∀Ψ ∈ Ker(T ∗) (3.11)

162 F. LANZARA

where Ker(T ∗) is the space formed by all the solutions of (3.10).Proof. If u is a solution of the BVP, then there exists (Φ; c) ∈ H such that(3.1) holds true (see Theorem 3.1) and

(f, Ψ) =(−1)l−1

4π(T (Φ; c), Ψ) = (−1)

l−1

4π〈(Φ; c), T ∗Ψ〉 = 0, ∀Ψ ∈ Ker(T ∗).

Conversely, suppose that conditions (3.11) hold true. Because of Theorem 3.1,we will prove Theorem 3.5 if we study the solvability of system (3.1). Since theoperator S is of regular type, there exists a reducing operator S ′ : [L2(∂A)]m →[L2(∂A)]m (see [15, p. 419], [21, p. 73]) such that S ′S = I + K where I is theidentity operator and K is a compact operator of the space [L2(∂A)]m.

Consider the operator T ′Ψ = (S ′Ψ; 0). T ′ is a continuous operator from[L2(∂A)]m → H and T ′T (Φ; c) = (Φ+KΦ+S ′Lc; 0). T ′ is a reducing operatorfor T because T ′T − I is a compact operator of the space H. Then (see [7,p. 138],[15, p. 420]) the Fredholm theorems hold and system (3.1) admits asolution if and only if conditions (3.11) are satisfied. ¤

4. An Index Formula for the BVP

Let χP be the index of the BVP, i.e., the difference between the dimensionof the space Ker(BV P ) of eigensolutions of the homogeneous problem associ-ated to BVP and the maximum number of linearly independent compatibilityconditions. By Theorem 3.5, we have

χP = dim Ker(BV P )− dim Ker(T ∗). (4.1)The homogeneous system associated to (2.6), i.e., (2.9), can be written as

S0Φ = 0. S0 is an operator of the space [L2(∂A)]m and

dim Ker(S0) = q l(l + 1)2

. (4.2)

Φ ∈ Ker(S0) if and only if v0[Φ] is a solution of (Pn). It follows that there existsc ∈ Cσn such that v0[Φ] = −p[c], where p[c] is given in (2.5). This shows thatu = v0[Φ] + p[c] is zero and then representation (2.4) of the solution of BVP isnot unique.

Theorem 4.1. In the hypotheses of the Representation Theorem 2.3, if ubelongs to V (A), there are uniquely determined (Φ⊥; d) ∈ Ker(S0)⊥×Cσn, suchthat

u = v0[Φ⊥] + p[d]. (4.3)

Proof. Let u be in V (A). For the Representation Theorem 2.3, there exists(Φ; c) ∈ H such that u = v0[Φ] + p[c]. If [L2(∂A)]m = Ker(S0) ⊕ Ker(S0)⊥, wehave Φ = Φ⊥⊕Φ0 with Φ⊥ ∈ Ker(S0)⊥ and Φ0 ∈ Ker(S0) uniquely determined.Then u = v0[Φ⊥]+v0[Φ0]+p[c]. Since Φ0 ∈ Ker(S0), v0[Φ0] is a solution of (Pn)and there exists b ∈ Cσn such that v0[Φ0] = p[b]. Hence follows the validity of(4.3) by assuming d = b + c.


Suppose that there exists (Ψ; d) 6= (0, 0), Ψ ∈ Ker(S0)⊥, such that, in A,v0[Ψ] + p[d] = 0. Then v0[Ψ] is a solution of (Pn). It follows that Ψ ∈ Ker(S0).Since Ψ ∈ Ker(S0) ∩Ker(S0)⊥, we have Ψ ≡ 0. Then d = 0. ¤

Let T̃ = T |Ker(S0)⊥×Cσn .Theorems 3.1 and 4.1 lead to a correspondence between the solutions of

T̃ (Φ; d) = 0. (4.4)and the solution of the homogeneous problem associated to BVP. Specifically,if u ∈ Ker(BV P ), then there exists (Φ; d) ∈ Ker(S0)⊥ × Cσn such that u =v0[Φ] + p[d] and T̃ (Φ; d) = 0. Conversely, if (Φ; d) ∈ Ker(S0)⊥ × Cσn satisfiesthe equation (4.4) then v0[Φ] + p[d] belongs to Ker(BV P ).

Moreover, if Ker(T̃ ) denotes the space of eigensolutions of (4.4), thenProposition 4.1. We have

dim Ker(BV P ) = dim Ker(T̃ ). (4.5)Proof. Let u(1), . . . , u(s) ∈ Ker(BV P ) be linearly independent. By Theorem 4.1there exist {(Φ(i); d(i))} ∈ Ker(S0)⊥×Cσn such that T̃ (Φ(i); d(i)) = 0 and u(i) =v0[Φ(i)] + p[d(i)], i = 1, . . . , s. If {c1, . . . , cs} are complex constants such that

s∑i=1

ciΦ(i) = 0,

s∑i=1

cid(i) = 0,

we deduce thats∑

i=1

ciu(i) = v0

[ s∑i=1

ciΦ(i)

]+ p

[ s∑i=1

cid(i)

]= 0

which implies c1 = · · · = cs = 0. Hence dim Ker(BV P ) 6 dim Ker(T̃ ).Conversely, if {(Φ(i); d(i))}i=1,...,s ∈ Ker(S0)⊥ × Cσn are linearly independent

solutions of (4.4), by assuming u(i) = v0[Φ(i)]+p[d(i)] we have u(i) ∈ Ker(BV P ).Therefore, if

s∑i=1

ciu(i) = 0, we deduce that

v0[ s∑

i=1

ciΦ(i)

]= p

[−

s∑i=1

cid(i)

]. (4.6)

Hence the function on the left-hand side of (4.6) is a solution of (Pn) i.e.s∑

i=1

ciΦ(i) ∈ Ker(S0). On the other hand, we have

s∑i=1

ciΦ(i) ∈ Ker(S0)⊥. Then

s∑i=1

ciΦ(i) = 0. From (4.6) we deduce that

s∑i=1

cid(i) = 0 and then c1 = · · · = cs =

0. Hence dim Ker(BV P ) > dim Ker(T̃ ). ¤From (4.1) and Proposition (4.5) we deduce that:

χP = dim Ker(T̃ )− dim Ker(T ∗).

164 F. LANZARA

Proposition 4.2. We have

dim Ker(T ) = dim Ker(T̃ ) + ql(l + 1)2

. (4.7)

Proof. Let s = dim Ker(T̃ ). Denote by {(Φ(i); d(i))}i=1,...,s a basis for Ker(T̃ ).Since Ker(T̃ ) ⊂ Ker(T ), let us assume {(Ψ(h); c(h))}h=1,...,r ⊂ Ker(T ) such that{(Φ(i); d(i))}i=1,...,s ∪ {(Ψ(h); c(h))}h=1,...,r is a basis for Ker(T ).

Now we will prove that r = ql(l+1)2

. Suppose that r > ql(l+1)2

. For a fixed h =

1, . . . , ql(l+1)2

+ 1, consider the function (Ψ(h); c(h)) and the corresponding u(h) =

v0[Ψ(h)] + p[c(h)]. Since u(h) ∈ V (A) and T (Ψ(h); c(h)) = 0, u(h) ∈ Ker(BV P )and there exists (χ(h); b(h)) ∈ Ker(S0)⊥ × Cσn such that u(h) = v0[χ(h)] + p[b(h)]with (χ(h); b(h)) ∈ Ker(T̃ ). It follows that

(χ(h); b(h)) =s∑

i=1

λ(h)i (Φ

(i); d(i)), {λ(h)1 , . . . , λ(h)s } 6= {0, . . . , 0}.

Then

v0[Ψ(h)] + p[c(h)] = v0[χ(h)] + p[b(h)] =s∑

i=1

λ(h)i v

0[Φ(i)] +s∑

i=1

λ(h)i p[d

(i)]

and therefore

p

[ s∑i=1

λ(h)i d

(i) − c(h)]

= v0[Ψ(h) −

s∑i=1

λ(h)i Φ

(i)

]. (4.8)

The function on the left-hand side of (4.8) is a solution of (Pn) and

Ψ(h) −s∑

i=1

λ(h)i Φ

(i) ∈ Ker(S0), h = 1, . . . , ql(l + 1)2

+ 1.

In view of (4.2), there exists µ = {µ1, . . . , µ ql(l+1)2

+1} 6= 0 such that

ql(l+1)2

+1∑

h=1

µh

(Ψ(h) −

s∑i=1

λ(h)i Φ

(i)

)= 0. (4.9)

(4.8) and (4.9) lead to

ql(l+1)2

+1∑

h=1

µh

(c(h) −

s∑i=1

λ(h)i d

(i)

)= 0. (4.10)

From (4.9) and (4.10) we deduce that {(Ψ(j); c(j))}j=1,...,

ql(l+1)2

+1and

{(Φ(i); d(i))}i=1,...,s are linearly dependent. This is impossible. It follows thatr 6 ql(l+1)

2.

Let {Ψ(h)}h=1,...,

ql(l+1)2

∈ Ker(S0) be linearly independent functions. Sincev0[Ψ(h)] is solution of (Pn), there exist b

(h) ∈ Cσn such that v0[Ψ(h)] = −p[b(h)].This implies that (Ψ(h); b(h)) ∈ Ker(T ), h = 1, . . . , ql(l+1)

2.


We will prove that (Ψ(h); b(h))h=1,...,

ql(l+1)2

and (Φ(j); d(j))j=1,...,s are linearly

independent functions in Ker(T ). Indeed, suppose thats∑

j=1

λj(Φ(j); d(j)) +

ql(l+1)2∑

h=1

µh(Ψ(h); b(h)) = 0

Since Ψ(h) ∈ Ker(S0) and Φ(j) ∈ Ker(S0)⊥ we deduce thats∑

j=1

λjΦ(j) =

ql(l+1)2∑

h=1

µhΨ(h) = 0,

which implies µh = 0, h = 1, . . . ,ql(l+1)

2. Then

s∑j=1

λj(Φ(j); d(j)) = 0

which implies λj = 0, j = 1, . . . , s, because {(Φ(j); d(j))}j=1,...,s ⊂ Ker(T̃ ) arelinearly independent.

As a consequence we have that

dim Ker(T ) > s + ql(l + 1)2

.

Thus r > ql(l + 1)2

and the theorem is proved. ¤

From (4.1), (4.5) and (4.7) we deduce that

χP = χT − ql(l + 1)2

, (4.11)

where χT denotes the index of the operator T , i.e. χT = dim Ker(T ) −dim Ker(T ∗) (see [21, p. 63]).

Let us denote by T1 and T2 the following operators of the space H →[L2(∂A)]m:

T1(Φ; c) = SΦ; T2(Φ; c) = Lc.Then T = T1 + T2. Since T2 is a compact operator, we have (see [15, p. 118])

χT = χT1 (4.12)

The following two equations

T1(Φ; c) = f, f ∈ [L2(∂A)]m, (4.13)and

SΦ = f, f ∈ [L2(∂A)]m, (4.14)are equivalent in the sense that they are both solvable or both unsolvable. SinceS is an operator of regular type, the necessary and sufficient conditions for thesolvability of equation (4.14), and hence of (4.13), are (f, Ψ) = 0, ∀Ψ ∈ Ker(S∗).Hence

dim Ker(S∗) = dim Ker(T ∗1 ). (4.15)

166 F. LANZARA

Set p = dim Ker(S). Let {Φ(i)}i=1,...,p be a basis for Ker(S).Then {(Φ(i); 0)}i=1,...,p belong to Ker(T1) and, together with

(0; 1, 0, . . . , 0︸︷︷︸σn−pla

), (0; 0, 1, . . . , 0︸︷︷︸σn−pla

), . . . , (0; 0, 0, . . . , 1︸︷︷︸σn−pla

) ,

form a basis for Ker(T1). Thusdim Ker(T1) = dim Ker(S) + σn. (4.16)

From (4.12), (4.15) and (4.16) we deduce that

χT = χS + σn. (4.17)

A final form for χP follows from (4.17), (4.11) and Theorem 2.2:

Theorem 4.2. The index of the BVP is given by

χP = χS + 2lnq.

Since S is of regular type, its index is given by the Muskhelishvili’s formula

χS =1

2πi

[log

det (αµ,σj(z)− βµ,σj(z))det (αµ,σj(z) + βµ,σj(z))

]

+∂A

,

where [ ]+∂A denotes the jump of the function between brackets after a coun-terclockwise tour along ∂A. This makes χP explicitly computable.

References

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(Received 13.07.2006)

Author’s address:

Dipartimento di MatematicaUniversità degli studi di Roma “La Sapienza”,Piazzale Aldo Moro 2, 00185 RomaItalyE-mail: [email protected]

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ON BVPS FOR STRONGLY ELLIPTIC SYSTEMS WITH HIGHER ORDER BOUNDARY CONDITIONS · 2007. 2. 23. · Representation (1.3) can also be used for solving boundary value problems with boundary