0

On substituting u(x) = ax, this reduces to »ax = ax/3, and so » = 1/3. All

other eigenvalues are zero. Our inhomogeneous equation was of the form

(1 ’ µK)u = f (9.59)

and the operator (1’µK) has an in¬nite set of eigenfunctions with eigenvalue

1, and a single eigenfunction, u0 (x) = x, with eigenvalue (1 ’ µ/3). The

eigenvalue becomes zero, and hence the inverse ceases to exist, when µ = 3.

A solution to the problem (1’µK)u = f may still exist even when µ = 3.

But now, applying the Fredholm alternative, we see that f must satisfy the

condition that it be orthogonal to all solutions of (1 ’ µK) † v = 0. Since our

kernel is Hermitian, this means that f must be orthogonal to the zero mode

u0 (x) = x. For the case of µ = 3, the equation is

1

u(x) = f (x) + 3 xyu(y) dy, (9.60)

0

1

and to have a solution f must obey 0 yf (y) dy = 0. We again set u =

f (x) + ax, and ¬nd

1 1

y 2 dy,

a=3 yf (y) dy + a3 (9.61)

0 0

but now this reduces to a = a. The general solution is therefore

u = f (x) + ax (9.62)

with a arbitrary.

266 CHAPTER 9. INTEGRAL EQUATIONS

9.5 Singular Integral Equations

Equations involving principal-part integrals, such as

P 1

1

•(x) dx = f (y), (9.63)

x’y

π ’1

in which f is known and we are to ¬nd •, are called singular integral equa-

tions. Their solution depends on what conditions are imposed on the un-

known function •(x) at the endpoints of the integration region. We will

1

consider only the simplest examples here.

9.5.1 Solution via Tchebychef Polynomials

Recall the de¬nition of the Tchebychef polynomials from chapter 2. We set

Tn (x) = cos(n cos’1 x), (9.64)

sin(n cos’1 x) 1

Un’1 (x) = = Tn (x). (9.65)

sin(cos’1 x) n

These are the Tchebychef Polynomials of the ¬rst and second kind, respec-

tively. The orthogonality of the functions cos nθ and sin nθ over the interval

[0, π] translates into

1

1

√ n, m ≥ 0,

Tn (x) Tm (x) dx = hn δnm , (9.66)

1 ’ x2

’1

where h0 = π, hn = π/2, n > 0, and

√ π

1

1 ’ x2 Un’1 (x) Um’1 (x) dx = δnm , n, m > 0. (9.67)

2

’1

Either of the sets {Tn (x)} and {Un (x)} are complete, and any L2 function

on [’1, 1] can be expanded in terms of them.

We also have the identities

1 1

1

√

P dx = 0, (9.68)

1 ’ x2 x ’ y

’1

1 1

1

√

P Tn (x) dx = π Un’1 (y), n > 0, (9.69)

x’y

1 ’ x2

’1

1

The classic text is N. I. Muskhelishvili Singular Integral Equations.

9.5. SINGULAR INTEGRAL EQUATIONS 267

and

√ 1

1

1 ’ x2 Un’1 (x)

dx = ’π Tn (y).

P (9.70)

x’y

’1

These are equivalent to the trigonometric integrals

cos nθ sin nφ

π

P dθ = π , (9.71)

cos θ ’ cos φ sin φ

0

and

sin θ sin nθ

π

dθ = ’π cos nφ,

P (9.72)

0 cos θ ’ cos φ

respectively. We will motivate and derive these formul¦ at the end of this

section.

From these principal-part integrals we can solve the integral equation

P 1

1

y ∈ [’1, 1],

•(x) dx = f (y), (9.73)

x’y

π ’1

for • in terms of f , subject to the condition that • be bounded at x = ±1.

We will see that no solution exists unless f satis¬es the condition

1

1

√ f (x) dx = 0, (9.74)

1 ’ x2

’1

but if f does satisfy this condition then the solution is

√

1 ’ y2 1 1

1

√

•(y) = ’ P f (x) dx. (9.75)

x’y

π 2

1’x

’1

To understand why this is the solution, and why there is a condition on f ,

expand

∞

f (x) = bn Tn (x). (9.76)

n=1

Here, the condition on f translates into the absence of a term involving

T0 ≡ 1 in the expansion. Then,

√ ∞

x2

1’