A solution method for the linearChandrasekhar equation

Elias Wegert and Lothar von Wolfersdorf

Dedicated to Professor J.Albrecht on the occasion of his 80th birthday

AMS subject classification: Primary 45E10, 45F15; Secondary 45G05, 65R20, 30E25

Keywords: Chandrasekhar equation, H-equation, Fredholm theory, Riemann-Hilbert prob-lem, Wiener-Hopf method

Abstract. The paper is devoted to the linearized H-equation of Chandrasekhar and Am-barzumyan. A Stieltjes-type transform reduces the equation to a boundary value problemfor holomorphic functions in the upper half-plane which is solved in closed form. Additionalconditions ensure that the solutions Φ extend holomorphically to the lower half-plane slitalong a straight line segment. The solutions of the original problem are then determinedfrom the boundary values of Φ on this slit.The approach gives necessary and sufficient conditions for Fredholmness and describes allFredholm parameters in terms of zeros of two functions 1−K and G associated with thekernel and the right-hand side of the equation. Explicit formulas for the complete set ofsolutions are presented.

Die Arbeit ist der linearisierten H-Gleichung von Chandrasekhar und Ambarzumyangewidmet. Die Gleichung wird mit Hilfe einer modifizierten Stieltjes-Transformation aufein Randwertproblem fur holomorphe Funktionen in der oberen Halbebene zuruckgefuhrtdas in geschlossener Form gelost wird. Unter zusatzlichen Bedingungen konnen die Lo-sungen holomorph in die langs einer Strecke aufgeschnittene untere Halbebene fortgesetztwerden. Die Losungen Φ des Ausgangsproblems werden dann aus den Randwerten von Φlangs des Schlitzes bestimmt. Der Zugang liefert notwendige und hinreichende Bedingun-gen dafur, dass der Operator Fredholmsch ist und charakterisiert alle Fredholmparametermit Hilfe der Nullstellen zweier Funktionen 1−K und G, die dem Kern und der rechtenSeite der Gleichung zugeordnet sind. Es werden explizite Darstellungen fur die vollstandigeLosungsmenge angegeben.

1

1 Introduction

One of the basic equations in the physics of radiation transfer and neutron transportis the so-called H-equation,

H(x)

(1− x

∫ 1

0

f(s) H(s)

s + xds

)= 1, 0 < x < 1 (1)

where f is a given function and H has to be determined. Introduced by V.A. Ambar-zumyan in 1941 and S. Chandrasekhar in 1947 as a tool for studying the radiativetransfer in a semi-infinite stellar atmosphere (see [4]), the equation has proved usefulin a number of applications (see especially Chandrasekhar [5], Busbridge [3], andthe overview in [19]). In 1966 M.G.Krein [14] gave a complete analytical treatmentof the general H-equation; a simplified solution method is proposed in [19].For practical purposes equation (1) is usually solved numerically by various iter-ation methods, see, for instance, the papers by G.A.Hively [10], C.T.Kelly [11],R.W.Legget [12, 13], and S. Seikkala [17, 18]. An interesting approximation methodhas been presented by J. Albrecht in his lecture at the GAMM conference in 1989[1]. Writing equation (1) as an appropriate fixed point equation, Albrecht obtainsmonotone inclusions of the solution.In the development of numerical methods for solving operator equations the H-equation (1) serves as a general test example. With respect to gradient methods asolution of the linearized equation

h(x)

(1− x

∫ 1

0

f(s) H0(s)

s + x

)ds− xH0(x)

∫ 1

0

f(s) h(s)

s + xds = g0(x) (2)

for the variation h of a solution H0 is of some interest. The operator on the lefthand-side represents the Frechet derivative of the non-linear operator in equation(1). Applying the usual gradient method to equation (1) may be considered as acorresponding symmetric mixed iteration method of Buckner-Wiarda type, usingsolutions of linear equations of form (2).Equation (2) has also relations to the theory of linear singular integral equationswith fixed singularities [6]. To the best of our knowledge, no analytic treatmentof this equation has been given in the literature, though it is certainly one of thesimplest practical relevant examples of an equation of the third kind which can besolved in closed form.The substitutions ϕ := f h, k := f H0, and g := f g0 transform equation (2) to

Aϕ(x) := ϕ(x)− x

∫ 1

0

ϕ(x) k(s) + k(x) ϕ(s)

s + xds = g(x), 0 < x < 1 (3)

2

which can be considered as an integral equation of the third kind with the coefficient

c(x) := 1− x

∫ 1

0

k(s)

s + xds (4)

and a kernel with peculiar behaviour at x = 0. A general class of related singularintegral equations with fixed singularity of the second kind has been investigatedby R.Duduchava in Section 8 of [6]. Note that we have a different situation here,because of the factor x in front of the integral (3).In the present paper we assume that the kernel k, the right-hand side g, and thesolution ϕ are real valued functions in the Lebesgue space Lr(0, 1) for some r with1 < r < ∞, and develop a solution method and all details of the Fredholm theory forthe linear Chandrasekhar operator A given by (3). The main result is the following.

Theorem 1. Let k ∈ Lr(0, 1) with 1 < r < ∞. Then the linear Chandrasekharoperator A given by (3) is bounded in Lr(0, 1) and has the following properties.

(i) The operator A is Fredholm if and only if the function 1−K with K definedby

K(z) := −iz

∫ 1

0

k(s)

s− i zds, z ∈ C \ [−i, 0] (5)

has no zeros on the straight line segment (0, i].

(ii) If A is Fredholm its index is zero and the dimension of its kernel is

dim ker A = κ1 +κ0

2+

⌊κ∞ + 1

2

⌋. (6)

Here κ0 and κ∞ denote the total number of zeros of 1−K on the real axis andat infinity, respectively, κ1 is the total number of common zeros of 1 − K(z)and 1−K(z) in C+ \ (0, i], and bxc stands for the integer part of x.

First of all we mention that it follows from the boundedness of x/(s+x) on (0, 1)×(0, 1) and Holder’s inequality that A is bounded in Lr(0, 1) with ‖A‖ ≤ 1 + 2 ‖k‖r,and that A depends continously in the operator norm on the kernel k ∈ Lr(0, 1).Before we go into the details we give an outline of the approach. In the first step aStieltjes-type transform

Tϕ(x) := x2

∫ 1

0

ϕ(s)

s2 + x2ds (7)

3

is applied which reduces (3) to a boundary problem of Riemann-Hilbert type forholomorphic functions in the upper half plane. Under certain explicit conditions thesolutions Φ admit a holomorphic continuation from C+ onto the Riemann sphere Ccut along the segment [−i, 0]. In the final step the solutions ϕ of the original problemare determined by Plemelj-Sokhotsky’s formula from the limits of Φ on both sidesof the cut [−i, 0]. The structure of the solution set depends on the index of thetransmission problem and on the location of zeros of the characteristic function1−K and of G.The plan of the paper is as follows. In Section 2 the integral equation (3) is re-duced to a Riemann-Hilbert problem on the upper half plane, which is solved inSection 3. In Section 4 the solutions of the Riemann-Hilbert problem are continuedholomorphically to the lower half-plane and in Section 5 the extended functions arerepresented by Cauchy integrals. The Fredholm properties of equation (3) are thenstudied in Section 6. Finally, in Section 7, an outlook to the linear Chandrasekharequation on the half-line R+ and to a system of two equations is given.

2 Transformation to Riemann-Hilbert Problem

In order to realize this programme we associate with the right-hand side g and thesolution ϕ the functions G and Φ, defined by

G(z) := z2

∫ 1

0

g(s)

s2 + z2ds, z ∈ C \ [−i, i], (8)

Φ(z) := −iz

∫ 1

0

ϕ(s)

s− i zds, z ∈ C \ [−i, 0]. (9)

The functions K, G and Φ are holomorphic in their domains of definition and havefinite limits at infinity,

Φ(∞) :=

∫ 1

0

ϕ(s) ds, K(∞) :=

∫ 1

0

k(s) ds, G(∞) :=

∫ 1

0

g(s) ds.

Since k, g and ϕ are real, K, G and Φ obey the symmetry properties

Φ(−z) = Φ(z), K(−z) = K(z), G(−z) = G(z) = G(z).

In particular all functions are real on the imaginary axis iR and G is also real onthe real axis R.

4

In order to study the continuity properties of K,G and Φ we give the followingdefinition.A function f defined on a subset X of C with ∞ ∈ X is said to be globally Holdercontinuous with exponent ν if it has a finite limit f(∞) at infinity and there existsa constant C such that for all finite x, y ∈ X

|f(x)− f(y)| ≤ C |x− y|ν , |f(x)− f(∞)| ≤ C |x|−ν . (10)

The space of all such functions is denoted by Cν(X).

Lemma 1. Let k, g ∈ Lr(0, 1) with r > 1 and 0 < ν < 1− 1/r. If K, G and Φ areextended by zero at z = 0, then K, Φ ∈ Cν(C+) and G ∈ Cν(R).

Proof. 1. For all real numbers u, v and 0 < ν ≤ 1 the elementary inequalities

|u|2−2ν ≤ 1 + u2, |u− v|2−2ν ≤ 2 (1 + u2)(1 + v2)

hold. The first one implies that for real x and s > 0

x2

s2 + x2=|x|2−2ν s2ν

s2 + x2

|x|2ν

s2ν=

|x/s|2−2ν

1 + (x/s)2

|x|2ν

s2ν≤ |x|2ν

s2ν

and the second one yields that

s2(x− y)2

(s2 + x2)(s2 + y2)=

|x/s− y/s|2−2ν

(1 + (x/s)2)(1 + (y/s)2)

|x− y|2ν

s2ν≤ 2

|x− y|2ν

s2ν

for all real x, y and s > 0. If z = a + ib with b ≥ 0 then

∣∣∣∣z

s− iz

∣∣∣∣2

=a2 + b2

(s + b)2 + a2≤ a2

s2 + a2+

b2

s2 + b2≤ |a|2ν

s2ν+|b|2ν

s2ν≤ 2

|z|2ν

s2ν. (11)

Similarly, for x = a + ib, y = c + id with b, d ≥ 0 and s > 0,

∣∣∣∣s(x− y)

(s− ix)(s− iy)

∣∣∣∣2

≤ 4|x− y|2ν

s2ν. (12)

2. For 0 < ν < 1 − 1/r the function s−ν belongs to Lr/(r−1)(0, 1) and then, byestimate (11) and Holder’s inequality,

|K(z)| ≤∫ 1

0

∣∣∣∣s

s− iz

∣∣∣∣ |k(s)| ds ≤ 2 |z|ν∫ 1

0

|k(s)|sν

ds ≤ C |z|ν

5

for all z in C+. Using the second estimate (12) we obtain analogously for x, y in C+

|K(x)−K(y)| ≤∫ 1

0

∣∣∣∣s(x− y)

(s− ix)(s− iy)

∣∣∣∣ |k(s)| ds ≤ C |x− y|ν .

The estimate at z = ∞ holds because the functions are holomorphic at infinity andthus Φ(z)− Φ(∞) = O(1/z).3. The proof for Φ is analogous and the result for G follows from the kernel decom-position

2z2

s2 + z2=

iz

s + iz− iz

s− iz.

Along the cut on the imaginary axis the functions K, G and Φ are in general dis-continuous and we introduce the one sided limits Φ± of Φ by

Φ±(−iy) := limx→±0

Φ(x− iy), y ∈ (0, 1).

In fact these limits also exist as nontangential limits. A similar definition is madefor K±(−iy) with y ∈ (0, 1) and G±(iy) with y ∈ (−1, 1). From Plemelj-Sokhotsky’sformula (see [16], Theorem 5.30, for instance) we get the following boundary values.

Lemma 2. If g, k, and ϕ ∈ Lr(0, 1) then for almost all y ∈ (0, 1)

G±(i y) = ±πi

2y g(y)− y2

∫ 1

0

g(s)

s2 − y2ds

G±(−i y) = ∓πi

2y g(y)− y2

∫ 1

0

g(s)

s2 − y2ds

K±(−i y) = ∓πi y k(y) − y

∫ 1

0

k(s)

s− yds

Φ±(−i y) = ∓πi y ϕ(y) − y

∫ 1

0

ϕ(s)

s− yds,

where the integrals on the right-hand sides are Cauchy principal value integrals.

Corollary 1. Under the assumptions of Lemma 2 we have the following jump rela-tions for almost all y ∈ (0, 1),

G−(−i y) −G+(−i y) = πi y g(y)

K−(−i y)−K+(−i y) = 2πi y k(y)

Φ−(−i y) − Φ+(−i y) = 2πi y ϕ(y).

6

In order to apply the Stieltjes-type transform T defined in (7) to equation (3), weremark that for all x ∈ R

Φ(x) = x2

∫ 1

0

ϕ(s)

s2 + x2ds− i x

∫ 1

0

s ϕ(s)

s2 + x2ds,

i.e., Tϕ(x) = Re Φ(x). The integral term on the left-hand side of (3) is transformedas follows:

T

(x

∫ 1

0

ϕ(x) k(s) + k(x) ϕ(s)

s + xds

)

= x2

∫ 1

0

s

s2 + x2

∫ 1

0

ϕ(s) k(t) + k(s) ϕ(t)

t + sdt ds

= x2

∫ 1

0

∫ 1

0

1

t + s

sϕ(s) k(t) + s ϕ(t) k(s)

s2 + x2dt ds

= x2

∫ 1

0

∫ 1

0

1

t + s

(s

s2 + x2+

t

t2 + x2

)ϕ(s) k(t) dt ds

= x2

∫ 1

0

∫ 1

0

x2 + st

(s2 + x2)(t2 + x2)ϕ(s) k(t) dt ds

= x2

∫ 1

0

t k(t)

t2 + x2dt ·

∫ 1

0

s ϕ(s)

s2 + x2ds + x4

∫ 1

0

k(t)

t2 + x2dt ·

∫ 1

0

ϕ(s)

s2 + x2ds

= x2 Re

(∫ 1

0

(t− ix) k(t)

t2 + x2dt ·

∫ 1

0

(s + ix) ϕ(s)

s2 + x2ds

)

= Re

(ix

∫ 1

0

k(t)

t + ixdt · (−ix)

∫ 1

0

ϕ(s)

s− ixds

)

= Re(K(x) Φ(x)

).

In this way equation (3) is reduced to a boundary value problem of Riemann-Hilberttype,

Re[(1−K) Φ

](x) = G(x) := Tg(x), x ∈ R (13)

for the holomorphic function Φ in the upper half plane. A solution Φ of (13) is saidto be symmetric if Φ(−z) = Φ(z) for all z ∈ C+.

Lemma 3. A function ϕ ∈ Lr(0, 1) is a solution of (3) if and only if its associatedfunction Φ defined by (9) is a symmetric solution of the Riemann-Hilbert problem(13) in the upper half-plane.

7

Proof. It only remains to show that the Stieltjes transform T is injective. If ϕ ∈Lr(0, 1) and Tϕ(x) = 0 on R, then the corresponding function Φ is holomorphicin C \ [−i, 0], Holder continuous on C+, and satisfies Re Φ(x) = Tϕ(x) = 0. Thisimplies that Φ(z) = i C on C+ and further C = 0 because Φ(0) = 0. By holomorphicextension Φ ≡ 0. Finally, by Corollary 1 we get ϕ ≡ 0.

Remark 1. Note that Lemma 3 does not claim that any symmetric solution Φ ofthe Riemann-Hilbert problem gives a solution ϕ of the Chandrasekhar equation; itis essential that the function Φ, which is associated with ϕ via (9), can be extendedto a holomorphic function in C \ [−i, 0] with (Φ−(iy)− Φ+(iy))/y ∈ Lr(0, 1).

3 Solution of the Riemann-Hilbert problem

Instead of quoting results from the classical theory (see [9], [15], for instance) we givea direct solution of the Riemann-Hilbert problem (13) based on the fact that 1−Kand G are holomorphic in C+. This leads to a special representation of the solutionand allows to treat the problem immediately also in some “degenerate” cases.Since some functions involved in the representation of the solution are not boundedwe introduce spaces Cν

γ of Holder continuous functions growing at infinity. For a

nonnegative integer γ and 0 < ν < 1 the space Cνγ (C+) consists of all functions

f which are defined on C+ such that (z + i)−γ f(z) ∈ Cν(C+). Analogously, f ∈Cν

γ (C−) if (z − i)−γ f(z) ∈ Cν(C−). Finally Cνγ (R), consists of all functions f on R

such that (x± i)−γ f(x) ∈ Cν(R).The explicit representation of the solutions to (13) and of its holomorphic continu-ation involve Cauchy-type operators S±γ defined on Cν

γ (R) by

S±γ f(z) :=1

2πi

∫ ∞

−∞

[(z + i

s + i

)γ

+

(z − i

s− i

)γ ]f(s)

s− zds, z ∈ C±. (14)

Here the positive signs are used if z ∈ C+ and the negative signs stand for z ∈ C−.The integral has to be understood in the sense of a Cauchy principal value at s = ∞.The functions S+

γ f and S−γ f are holomorphic in the upper and lower half-plane,respectively. Some relevant properties of S±γ are summarized in the next lemma.

Lemma 4. Let f ∈ Cνγ (R) with 0 < ν < 1 and γ ∈ Z+.

(i) The functions S+γ f and S−γ f extend continuously to functions in Cν

γ (C+) and

Cνγ (C−), respectively.

8

(ii) The boundary functions of S+γ f and S−γ f on R are given by

S+γ f(x) = Sγf(x) + f(x), S−γ f(x) = Sγf(x)− f(x), x ∈ R,

where Sγ is the singular integral of Cauchy type

Sγf(x) :=1

2πi

∫ ∞

−∞

[(x + i

s + i

)γ

+

(x− i

s− i

)γ ]f(s)

s− xds, x ∈ R. (15)

(iii) The operators Sγ satisfy the symmetry relation Sγf = −Sγf .

Note that the integral in (15) is a Cauchy principle value integral at s = x ands = ∞.

Proof. The operators S±0 are (apart from a factor of 2) the usual Cauchy integralsS+ and S− in the upper and lower half plane, respectively,

S±f(z) :=1

πi

∫ ∞

−∞

f(s)

s− zds, z ∈ C±, (16)

and S := S0 is the singular integral operator of Cauchy type on the real line. Sofor γ = 0 all assertions are standard results, see Muskhelishvili [15], for instance. Ifγ > 0 we write

S±γ f(z) =1

2(z + i)γS±f+(z) +

1

2(z − i)γS±f−(z),

where f±(s) := (s± i)−γf(s), and apply the result with γ = 0.

The following corollary is an immediate consequence of (ii) and (iii) and shows thatthe operator S+

γ gives a particular solution of the simplest Riemann-Hilbert problemRe F = f on the upper half-plane.

Corollary 2. If f ∈ Cνγ (R) is real, then Re S+

γ f(x) = f(x) on R.

In the special situation of the Riemann-Hilbert problem (13) the set of symmetricsolutions depends on the location of zeros and the behaviour of 1−K at infinity.First of all we remark that the total number of zeros of 1 −K in the closed upperhalf-plane C+ is always finite, because 1 −K is holomorphic at infinity, K(0) = 0,and K is Holder continuous on C+. In particular, at infinity 1−K is of finite orderz−γ with a non-negative integer γ.

9

Let z1, . . . , zn and x1, . . . , xm be the zeros of 1 − K in the upper half-plane C+

and on the positive real axis R+, with corresponding multiplicities α1, . . . , αn andδ1, . . . , δm, respectively. Further we define the functions R and N by

R(z) :=m∏

j=1

(z2 − x2j)

δj ·n∏

j=1

(z − zj)αj(z − zj)

αj , (17)

N(z) :=m∏

j=1

(z2 − x2j)

δj(z2 + 1)−δj . (18)

As usual, all empty products are set to one. It is easy to see that these functionsare symmetric,

R(−z) = R(z) = R(z), N(−z) = N(z) = N(z).

Lemma 5. Let K and G be given by (5), (8) with k, g ∈ Lr(0, 1), 1 < r < ∞,denote by z1, . . . , zn ∈ C+ and x1, . . . , xm ∈ R+ the zeros of 1 − K in C+ and byα1, . . . , αn and δ1, . . . , δm the corresponding multiplicities, suppose that 1 −K is oforder −γ at infinity, and let

α :=n∑

j=1

αj, δ :=m∑

j=1

δj.

(i) The Riemann-Hilbert problem (13) has a symmetric solution Φ ∈ Cν(C+) with0 < ν < 1−1/r if and only if the right-hand side G has zeros of order not lessthan δj at xj for j = 1, . . . , m and a zero of order not less than γ at infinity.

(ii) Assume that G satisfies the conditions of (i) and let R and N be defined by(17) and (18), respectively. Then the general symmetric solution of (13) inCν(C+) is given on C+ by

Φ ≡ Φ+ :=1−K

NS+

γ

(NG

|1−K|2)

+ i P1−K

R, (19)

where P is an arbitrary real odd polynomial of degree at most 2α + 2δ + γ.

(iii) All solutions vanish at the origin.

Proof. 1. The necessity of the conditions given in (i) follows immediately from theboundary relation (13).

10

2. Suppose that G satisfies the solvability conditions. In order to prove that thenthe function Φ+ defined by (19) with P ≡ 0 is a particular solution of (13) weremark that the function GN/|1−K|2 is real on R and belongs to Cν

γ (R). Then, byCorollary 2, we have on R

Re [ (1−K)Φ+] =|1−K|2

NRe S+

γ

(GN

|1−K|2)

=|1−K|2

N

GN

|1−K|2 = G.

It follows from Lemma 4 (i) that S+γ (GN/|1 − K|2) ∈ Cν

γ (C+) and then zero-polecancellation and the asymptotic behaviour of 1 −K and N at infinity ensure thatΦ+ ∈ Cν(C+). A straightforward computation finally shows that Φ+ satisfies thesymmetry condition Φ+(−z) = Φ+(z).

3. It remains to find the solutions of the homogeneous equation. Since the zerosof 1 − K are cancelled by the zeros of R, for any such solution Φ the functionQ := Φ R/(1 − K) is holomorphic on C+ and has vanishing real part on R. BySchwarz’ reflection principle it extends holomorphically onto C− to an entire functiongrowing at most like z2α+2δ+γ at infinity. By Liouville’s theorem Q must be apolynomial of degree not exceeding 2α + 2δ + γ. The boundary condition Re Q = 0implies that Q = iP with a real polynomial P , and by virtue of the symmetryproperties of R, K and Φ, the polynomial P must be odd.Conversely, it is easy to verify that any function Φ := i P (1 − K)/R with such apolynomial P is a symmetric solution of the homogeneous problem.

4. Finally, Φ(0) = Φ(0), K(0) = 0, and Re [(1 − K)Φ ](0) = G(0) = 0 yield thatΦ(0) = 0.

Remark 2. Because G is real and even on R and holomorphic at infinity, thesolvability conditions in (i) are equivalent to

dkG

dxk(xj) = 0, j = 1, . . . , m, k = 0, . . . , δj − 1, (20)

limx→∞

|x|γ−1 G(x) = 0. (21)

The conditions (20) are linearly independent over R, whereas (21) consists only ofb(γ + 1)/2c linearly independent conditions since G has polynomial behaviour in1/z2 at infinity. All conditions in (i) can be expressed in terms of continuous linearfunctionals (integrals) acting on g ∈ Lr(0, 1) in the representation (8) of G. Forinstance, (21) is equivalent to

∫ 1

0

s2kg(s) ds = 0, k = 0, 1, . . . , b(γ − 1)/2c.

11

Remark 3. Since P is real and odd, the homogeneous problem with G ≡ 0 hasα + δ + b(γ + 1)/2c real linearly independent solutions.

4 Holomorphic continuation

In this section we investigate under which conditions the solutions Φ+ of the Rie-mann-Hilbert problem admit a holomorphic extension to the slit sphere C \ [−i, 0].

Lemma 6. Let K, G, N , P and R be as in Lemma 5 and assume that G satisfiesthe solvability conditions of Lemma 5 (i). Then the solution Φ+ defined in (19) hasa holomorphic extension to C \ [−i, 0] if and only if the meromorphic function

ΦP (z) :=2 G(z)

1−K(−z)+ iP (z)

1−K(z)

R(z), z ∈ C \ [−i, i], (22)

has no poles in C− \ [−i, 0). In this case the extension is given by

Φ−(z) :=1−K(z)

N(z)S−γ

(NG

|1−K|2)

(z) + ΦP (z), z ∈ C− \ [−i, 0). (23)

Proof. If G satisfies the solvability conditions of Lemma 5 (i) then the functionNG/|1 −K|2 belongs to Cν

γ (R+) and the first summand on the right-hand side of(23) is holomorphic in C− \ [−i, 0) and bounded near infinity. The boundary valuesof Φ− and Φ+ on the real line coincide, since by Lemma 4

Φ+(x)− Φ−(x) =2 (1−K(x))G(x)

|1−K(x)|2 − 2 G(x)

1−K(−x),

and K(−x) = K(x) for x ∈ R. So Φ+ and Φ− together define a meromorphicfunction on C \ [−i, i]. The point at infinity is a removable singularity, since Φ+

and and Φ− are bounded in a neighbourhood of z = ∞ in C+ and C−, respectively.Hence Φ− is always the unique meromorphic continuation of Φ+ onto C− \ [−i, 0).This extension is holomorphic if and only if ΦP has no poles in C− \ [−i, 0).

In the next step we formulate conditions on G and P which guarantee that ΦP hasno poles. All possible poles of ΦP are among the zeros z1, . . . , zn of 1−K in the upperhalf-plane. Depending on the location of zj we split the index set J = {1, . . . , n}

12

into four classes:

j ∈ J+ if Re zj > 0, Im zj > 0

j ∈ J− if Re zj < 0, Im zj > 0

j ∈ J0 if Re zj = 0, Im zj > 1

j ∈ J∗0 if Re zj = 0, 0 < Im zj ≤ 1.

In the following we assume that j ∈ J+∪J−∪J0 and denote by βj ≥ 0 the multiplicityof the zero zj of K1 := 1−K. Then zj is a zero of order βj of K1 := 1−K and a zero oforder αj of R and K2 defined by K2(z) := 1−K(−z), respectively. The function ΦP

is finite at zj if and only if the meromorphic function Q := RΦP = i K1P +2 GR/K2

has a zero of order not less than αj at zj. In the case αj ≤ βj the function K1 hasa zero of order not less than αj. Further R/K2 is holomorphic and different fromzero near zj. Therefore ΦP is finite at zj if and only if zj is a zero of G of order notless than αj.In the case αj > βj the function ΦP is finite at zj if and only if the functionQ0 := Q/(iK1) = P − 2i GR/(K1K2) has a zero of order not less than αj − βj at zj.This is fulfilled if and only if zj is a zero of order not less than βj of G and furtherdkQ0/dzk (zj) = 0 for k = 0, . . . , αj − βj − 1.Both cases can be summarized in the following αj conditions,

dkG

dzk(zj) = 0, k = 0, . . . , min{αj, βj} − 1 (24)

dkP

dzk(zj) = 2i

dk

dzk

(GR

M

)(zj), k = 0, . . . , max{0, αj − βj} − 1, (25)

where M is defined by

M(z) := (1−K(z))(1−K(−z)). (26)

Since for j ∈ J∗0 the points zj are outside the domain C− \ [−i, 0], a holomorphiccontinuation of Φ+ to C− \ [−i, 0] exists if and only if conditions (24) and (25) holdfor all j ∈ J+ ∪ J− ∪ J0.It remains to find a complete set of linearly independent conditions. Due to thesymmetries G(−z) = G(z) and Q(−z) = −Q(z) all conditions with j ∈ J− areequivalent to the corresponding conditions with j ∈ J+. Since G and iP are real on

13

the imaginary axis, for j ∈ J0 the conditions (24) and (25) reduce to

Re

[ik

dkG

dzk

](zj) = 0, k = 0, . . . , min{αj, βj} − 1 (27)

Re

[ik

dk

dzk

(i P + 2

GR

M

)](zj) = 0, k = 0, . . . , αj − βj − 1. (28)

Indeed, the set of conditions (24) with j ∈ J+ and (27) with j ∈ J0 is linearlyindependent on the space of functions G generated by g ∈ Lr(0, 1). This can be seenfrom the explicit integral representations of the corresponding linear functionals,which involve the linearly independent functions

1

s2 + z2j

,s2

(s2 + z2j)

k+1, j, k = 1, 2, . . . .

Also, for G ≡ 0, conditions (25) for j ∈ J+ together with (28) for j ∈ J0 arelinearly independent on the space of real odd polynomials of degree not exceeding2α−1. This follows by interpreting the conditions as data of a Hermite interpolationproblem for the polynomial P0 defined by P (z) = z P0(z

2).

5 Representation by Cauchy integrals

If G and P satisfy the solvability conditions of Lemma 5 and Lemma 6 the solutionΦ+ of the Riemann-Hilbert problem has a holomorphic extension Φ to C\ [−i, 0]. Itremains to investigate if Φ can be represented in the form (9). The following Lemmagives a criterion.

Lemma 7. Let G and P satisfy the solvability conditions of Lemma 5 and Lemma 6and let Φ be defined by (19) for z ∈ C+ and by (22), (23) for z ∈ C−.

(i) For 1 < p ≤ r the function Φ admits a representation (9) with ϕ ∈ Lp(0, 1) ifand only if

supx 6=0

∫ ∞

−∞

∣∣∣∣Φ(x + iy)

x + iy

∣∣∣∣p

dy < +∞. (29)

The corresponding density ϕ is then given by

ϕ(y) =g(y)

1−K(iy)− k(y)

[1

N(iy)S−γ

(NG

|1−K|2)

(−iy)− i P (iy)

R(iy)

](30)

almost everywhere on (0, 1).

14

(ii) If 1−K has no zeros on [0, i] then all functions Φ admit a representation (9)with ϕ ∈ Lr(0, 1).

Corollary 3. If Φ has a representation (9) with density ϕ in Lr(0, 1) then theright-hand side of (30) must belong to Lr(0, 1).

Remark 4. With some more effort it can be shown that for 1 < p < r condition(29) can be replaced by

supx>0

∫ 1+ε

0

|ΦP (x− iy)|p dy < +∞. (31)

where ΦP is defined by (22) and ε is an arbitrary positive number.

Proof. 1. We remark that (9) can be interpreted as Cauchy integral representationof Φ(z)/z in the left half-plane CL and in the right half-plane CR, respectively, withdensity ϕ on the imaginary axis. If ϕ is in Lp then Φ(z)/z belongs to the Hardyspaces Hp(CL) and Hp(CR) which is exactly the meaning of (29).

2. Conversely, if condition (29) is satisfied, then Φ(z)/z can be represented in theleft half-plane as Cauchy integral over its boundary function ϕ− ∈ Lp(R) defined byϕ−(y) := Φ−(iy)/(iy), where

Φ±(iy) := limx→±0

Φ(x + iy),

namely,

1

2πi

∫ ∞

−∞

ϕ−(y)

iy − zd(iy) =

{Φ(z)/z if Re z < 0

0 if Re z > 0

(see Theorem 5.19 in [16], for instance). Analogously, in the right half-plane,

1

2πi

∫ ∞

−∞

ϕ+(y)

iy − zd(iy) =

{0 if Re z < 0

−Φ(z)/z if Re z > 0

where ϕ+(y) := Φ+(iy)/(iy) also belongs to Lp. Subtracting these representations,the values of ϕ− and ϕ+ cancel outside the interval [−1, 0], which leads to

Φ(z) = − z

2πi

∫ 0

−1

ϕ+(y)− ϕ−(y)

y + izdy = −iz

∫ 1

0

ϕ(y)

y − izdy, Re z 6= 0,

15

where

ϕ(y) :=Φ−(−iy)− Φ+(−iy)

2πiy.

Finally, the density ϕ can be determined from the representation (23) of Φ in thelower half plane and the jump relations of Corollary 1. Using the symmetries of K,N , P and R we so obtain (30), which also yields that ϕ is real-valued.

3. In order to prove (ii) we show that in this case condition (29) is satisfied for1 < p < r. Because the involved functions are symmetric it suffices to considerx > 0.3.1. Since Φ+(0) = 0 and Φ+ is Holder continuous on C+ we get the estimate

|Φ+(z)/z| ≤ C min{|z|−1, |z|ν−1}, z ∈ C+ (32)

with 0 < ν < 1−1/r. Choosing ν sufficiently close to 1−1/r we obtain for all p < r

supx>0

∫ +∞

0

∣∣∣∣Φ(x + iy)

x + iy

∣∣∣∣p

dy < +∞. (33)

3.2. Recall that in the lower half-plane Φ = Φ0 + ΦP , where

Φ0(z) :=1−K(z)

N(z)S−γ

(NG

|1−K|2)

(z),

ΦP (z) := i P (z)1−K(z)

R(z)+

2 G(z)

1−K(−z).

We now choose a positive number ε so small that the compact rectangle

D1 := {x + i y : 0 ≤ x ≤ ε, −1− ε ≤ y ≤ 0}contains no zeros of N(z), R(z) and 1−K(−z), and set

D2 := {x + i y : 0 ≤ x ≤ ε, y ≤ −1− ε},D3 := {x + i y : x ≥ ε, y ≤ 0}.

The function Ψ := S−γ (NG/|1−K|2) belongs to Cνγ (C−) and vanishes at z = 0. In

D1 we have∣∣∣∣Φ0(z)

z

∣∣∣∣ ≤ C

∣∣∣∣Ψ(z)

z

∣∣∣∣ + C

∣∣∣∣K(z)

z

∣∣∣∣ ,

16

and since |Ψ(z)/z| ≤ C |z|ν−1 and K(z)/z ∈ Hr(CR) we get the estimate

sup0<x≤ε

∫ 0

−1−ε

∣∣∣∣Φ0(z)

z

∣∣∣∣p

dy < +∞. (34)

In D2 and D3 the function (1−K)/N is bounded and behaves like |z|−γ at infinity,which implies that |Φ0(z)/z| is bounded and |Φ0(z)/z| behaves like |1/z|, so that forall p < r

sup0<x≤ε

∫ 1−ε

−∞

∣∣∣∣Φ0(z)

z

∣∣∣∣p

dy < +∞, sup0<x≤ε

∫ 0

−∞

∣∣∣∣Φ0(z)

z

∣∣∣∣p

dy < +∞.

Together with (34) this gives

supx>0

∫ 0

−∞

∣∣∣∣Φ0(x + iy)

x + iy

∣∣∣∣p

dy < +∞. (35)

In order to estimate the corresponding integral over ΦP we remark that the functions1/(1−K(−z)), 1/R(z), and P (z)/z are bounded on D1 (note that P (0) = 0), hence

|ΦP (z)/z| ≤ C (1 + |K(z)|+ |G(z)/z|) , z ∈ D1. (36)

In D2 and D3 the function ΦP is bounded, such that for all p > 1

supx>ε

∫ 0

−∞

∣∣∣∣ΦP (z)

z

∣∣∣∣p

dy ≤ C

∫ ε

0

1

|ε + i y|p dy < +∞, (37)

sup0<x≤ε

∫ −1−ε

−∞

∣∣∣∣ΦP (z)

z

∣∣∣∣p

dy ≤ C

∫ ∞

1+ε

1

|ε + i y|p dy < +∞. (38)

Finally, G(z)/z and K(z)/z belong to Hr(CR) and thus we infer from (36) that forall p ≤ r

sup0<x≤ε

∫ 0

−1−ε

∣∣∣∣ΦP (z)

z

∣∣∣∣p

dy ≤ C sup0<x≤ε

∫ 0

−1−ε

(1 +

∣∣∣∣K(z)

z

∣∣∣∣r

+

∣∣∣∣G(z)

z

∣∣∣∣r)

dy < +∞.

Combining this estimate with (37) and (38) the estimate (29) follows for all p < r.Consequently Φ has the representation (9) with ϕ ∈ Lp(0, 1) given by (30). Since allfunctions involved in the representation (30) are Holder continuous, except g and kwhich are in Lr(0, 1), it follows that ϕ in fact belongs to Lr(0, 1).

17

6 Fredholm properties of the operator A

The following theorem describes the solvability of the linear Chandrasekhar equation(3) in the case where 1−K has no zeros on the critical segment [0, i].

Theorem 2. Let k ∈ Lr(0, 1) with 1 < r < ∞ and suppose that the characteristicfunction 1 − K has no zeros on the segment [0, i]. Let zj (j = 1, . . . , n) and xl

(l = 1, . . . , m) be the zeros of 1 − K in the upper half plane and on the positivereal axis, respectively, with corresponding multiplicities αj and δl. Further denote byβj ≥ 0 (j = 1, . . . , n) the multiplicities of the zeros zj of 1−K. Finally let −γ ≤ 0be the order of 1−K at infinity. Then the following assertions hold.

(i) The linear Chandrasekhar operator A : Lr(0, 1) → Lr(0, 1) defined by (3) isFredholm with index zero and the dimension of its kernel is equal to

dim ker A :=

⌊γ + 1

2

⌋+

n∑j=1

min{αj, βj}+m∑

l=1

δl. (39)

(ii) Let further G and M be defined by (8) and (26), respectively. A solution of thelinear Chandrasekhar equation (3) exists if and only if G satisfies the conditions(24) for all zeros zj with Re zj > 0, (27) for all zeros zj with Re zj = 0, (20)for all zeros xl ∈ R+, and (21) at infinity.

(iii) Let further R and N be defined by (17) and (18), respectively. If G satisfiesthe conditions of (ii), the general solution of (3) is given by (30) with any realodd polynomial P of degree not exceeding

δ := γ + 2n∑

j=1

αj + 2m∑

l=1

δl

satisfying the interpolation conditions (25) for all zeros zj with Re zj > 0 and(28) for all zeros zj with Re zj = 0.

Proof. 1. Let ϕ ∈ Lr(0, 1) be a solution of (3). Then Φ defined by (9) is holomorphicin C \ [−i, 0], belongs to Cν(C+) with 0 < ν < 1− 1/r and Φ(z)/z is in the Hardyspace Hr(CR) on the right half-plane CR.By Lemma 3 Φ is a symmetric solution of the Riemann-Hilbert problem (13). HenceG must satisfy the solvability conditions (20), (21) of Lemma 5 (i), and Φ on C+

has the form described in Lemma 5 (ii).

18

Since Φ extends holomorphically onto C\ [−i, 0], the function G and the polynomialP must satisfy the conditions (24), (25) (27), (28) of Lemma 6. Finally, by Lemma 7,the density ϕ in (9) is given by (30).

2. Conversely, let G satisfy the conditions (ii) of Theorem 2. Then there existsan odd polynomial P of degree not greater than δ which solves the interpolationconditions (25) and (28). For any such polynomial the function ϕ given by (30)belongs to Lr(0, 1) and the function Φ defined by (9) is holomorphic in C \ [−i, 0].Interpreting (9) as a Cauchy integral for Φ(z)/z we see that Φ and the right-handside of (23) with P ≡ 0 coincide on C− \ [−i, 0]. Since the right-hand side of (19)is the holomorphic continuation of the right-hand side of (23), Φ must by given by(19) on the upper half plane. So Φ is a symmetric solution of the Riemann-Hilbertproblem (13) which, by Lemma 3, is equivalent to the Chandrasekhar equation (3)for ϕ.

3. In order to prove (i) we just remark that all conditions on G and P can beexpressed in terms of bounded linear functionals. Further, from (30) with (25), (28)and Lemma 5 with (24), (27), respectively, we infer (after some lengthy computa-tion) that both the number of linearly independent (real) solutions of (3) and thenumber of linearly independent (real) solvability conditions on the right-hand sideg of equation (3) are given by (39).

Finally, we investigate what happens if 1−K has zeros on (0, i]. Though those zerosdo not disturb the holomorphic extension of Φ, they influence the representationof Φ(z) by (9) with density ϕ ∈ Lr(0, 1). By Lemma 7 such a representation (anda solution of equation (3)) exists if and only if Φ satisfies condition (29). If thiscondition is fulfilled depends exclusively on the behaviour of K, G, and P near thepoints −iyj where iyj are the zeros of 1−K on [0, i]. Though (29) gives a reasonablecriterion (see also Remark 4) for solvability of a concrete equation (3), it is ratherbad in the context of Fredholm theory.

Lemma 8. If 1−K has zeros on (0, i] then A is not Fredholm.

Proof. 1. In the first step we assume that 1 − K has a zero iy0 with 0 < y0 < 1.We prove that then the codimension of im A is infinite. In fact it follows fromCorollary 3 that a function g ∈ Lr(0, 1) can only belong to im A if there exists apolynomial P (having certain additional properties) such that ϕ defined by (30)belongs to Lr(0, 1), which is equivalent to

g(y)

1−K(iy)+ i k(y)

P (iy)

R(iy)∈ Lr(0, 1).

19

Since P (iy) can compensate the pole at y0 only for one value of g(y), this is certainlynot the case if g is continuous on [0, 1] with the exception of a jump at y0.

2. If i is the only zero of 1−K on the segment (0, i], we consider the operators Aλ

with kernel λk where λ > 1. Then the real-valued continuous function 1 − λK ispositive at z = 0 and negative at z = i and hence it must have a zero in (0, i). Bythe first step of the proof Aλ is not Fredholm for all λ > 1, which is only possible ifA itself is not Fredholm, because the mapping λ 7→ Aλ is continuous (see Section 1)and the set of Fredholm operators is open.

Remark 5. If 1−K has zeros iyj on (0, i) and k is sufficiently smooth, the integralequation (3) would be Fredholm in spaces of smooth functions on [0, 1]. Then thereoccur conditions at −iyj±0 on both sides of the slit [−i, 0], which leads to a negativeindex of the linear Chandrasekhar operator A. In this case the corresponding solv-ability conditions are point conditions and integral conditions with Cauchy integralsof g.The case where 1−K has zeros on (0, i) corresponds to the singular case of vanishingcoefficient c(x) at x = yj in equation (3) of third kind.

The statements of Theorem 1 now follow from Theorem 2 and Lemma 8. Note that∑αj is the number of zeros in the upper half plane counted with multiplicity, while∑δl is half of that number for zeros on the real line.

As a consequence of Theorem 1 we formulate the following simple criterion on Fred-holmness and invertibility.

Theorem 3. Let k ∈ Lr(0, 1) with 1 < r < ∞. Then the linear Chandrasekharoperator A is Fredholm in Lr(0, 1) if and only if

K(iy) ≡ y

∫ 1

0

k(s)

s + yds < 1 for all y ∈ [0, 1]. (40)

The operator is invertible in Lr(0, 1) if and only if (40) is satisfied and

|1−K(z)|+ |1−K(z)| 6= 0 for all z ∈ C \ [−i, i]. (41)

Remark 6. Sufficiency of condition (40) for Fredholmness of the operator A followsdirectly from (3), since x/(s+x) is a bounded function so that after division by thenon-vanishing continuous function c from (4) the integral operator in (3) is compactin Lr(0, 1).

20

Remark 7. Condition (40) corresponds to the condition

1 + iz

∫ 1

0

f(s)H(s)

s− izds

∣∣∣∣z=iy

6= 0

necessary for the boundedness of solutions to the nonlinear H-equation (1), see [19].

A simple perturbation argument shows that “generic” linear Chandrasekhar oper-ators are either invertible or not Fredholm. In particular A is not Fredholm if k islarge.

7 Outlook

The presented solution method can be extended to the integral equation

ϕ(x)−∫ ∞

0

ϕ(x)k(s) + k(x)ϕ(s)

s + xds = g(x), x > 0 (42)

on the positive half line R+, with real-valued summable functions k and g, by ap-plying the Stieltjes transform

Tϕ(x) =

∫ ∞

0

s ϕ(s)

s2 + x2ds.

Equation (42) belongs to a class of integral equations on the half-axis studied byF.D.Berkovic [2].Correspondingly, systems of equations

ϕ1(x)− x

∫ 1

0

ϕ1(x)k1(s) + k2(x)ϕ2(s)

s + xds = g1(x), 0 < x < 1

ϕ2(x)− x

∫ 1

0

ϕ2(x)k2(s) + k1(x)ϕ1(s)

s + xds = g2(x), 0 < x < 1

can be dealt with applying the complex-valued Stieltjes transform

T (ϕ1, ϕ2)(x) = −ix

∫ 1

0

ϕ1(s)

s− ixds + ix

∫ 1

0

ϕ2(s)

s + ixds

(see [20]). A related system of quadratic H-equations on R+ was introduced byN.B.Engibaryan and his coworkers [7, 8].

21

References

[1] J. Albrecht, Einschließung der Losung der Integralgleichung von Chandrase-khar, ZAMM (J.Appl.Math. Mech.) 70, 6, T588–590 (1990).

[2] F.D.Berkovic , On an integral equation on the half-axis (Russ.), Izvestiija VusovMatematika (Kazan) 1966, 1(50), 3–14.

[3] T.W.Busbridge, The Mathematics of Radiative Transfer. Cambridge UniversityPress, 1960.

[4] S. Chandrasekhar, The transfer of radiation in stellar atmospheres, Bull. Amer.Math. Soc. 53, 641–711 (1947).

[5] S. Chandrasekhar, Radiative Transfer, Clarendon Press, Oxford, 1950.

[6] R.Duduchava, Integral Equations with Fixed Singularities, Teubner, Leipzig,1979.

[7] N.B. Engibaryan and A.A.Aruntynyan, Integral Equations on the half-line withdifference kernels, and nonlinear functional equations, Math. USSR Sbornik,26, 35–58 (1975).

[8] N.B. Engibaryan and L.G. Arabadzhyan, Systems of Wiener-Hopf integral equa-tions, and nonlinear factorization equations, Math. USSR Sbornik, 52, 121–208(1985).

[9] F.D.Gakhov, Boundary Value Problems, Pergamon Press, Oxford, 1966.

[10] G.A.Hively, On a class of nonlinear integral equations arising in transport the-ory, SIAM J. Math. Anal. 9, 787–792 (1978).

[11] C.T.Kelly, Solutions of H-equation by iteration, SIAM J. Math. Anal. 10,844–849 (1979).

[12] R.W.Legget, A new approach to the H-equation of Chandrasekhar, SIAM J.Math. Anal. 7, 542–550 (1976).

[13] R.W.Legget, On certain nonlinear integral equations, J. Math. Anal. Appl. 57,462–468 (1977).

22

[14] M.G.Krein, On nonlinear integral equations which play a role in the theoryof Wiener-Hopf equations I,II. In: M.G. Krein, Topics in Differential and Inte-gral Equations and Operator Theory, Operator Theory and Applications, Vol. 7(I.Gohberg Ed.), Birkhauser, Basel 1983, pp. 173–242.

[15] N.I.Muskhelishvili, Singular Integral Equations, Noordhoff, Groningen, 1953.

[16] M.Rosenblum and J.Rovnyak, Topics in Hardy Classes and Univalent Func-tions. Birkauser, Basel 1994.

[17] S. Seikkala, Iterations for an H-equation, Acta Univ. Ouluen A 144, 1–17(1983).

[18] S. Seikkala, Solutions by a series of the Chandrasekhar H-equation, Proceed.Summer School Num. Anal., Jyvaskyla, 1985, pp. 291–298.

[19] L. von Wolfersdorf, Analytische Losungen der Auto-, Kreuz- und Tripel-Korre-lationsgleichungen und verwandter Integral- und Integrodifferentialgleichungen,Sitzber. Sachs. Akad. Wiss. Leipzig, Math.-Nat. Kl., Band 129, Heft 6, 2004.

[20] L. von Wolfersdorf, A class of linear integral equations and systems with sumand difference kernel, ZAA (J. Anal. Appl.), 22, 647–687 (2003).

23