Now, Laplace transforms have the property that
d
pLF = L F, (9.41)
dx
as may be seen by an integration by parts in the de¬nition. Using this, and
depending on whether we put the p next to f or outside the parenthesis, we
conclude that the solution of Abel™s equation can be written in two equivalent
ways:
1d 1 1 1
x x
√ √
u(x) = f (y) dy = f (y) dy. (9.42)
x’y x’y
π dx π
0 0
Proving the equality of these two expressions was a problem we set ourselves
in chapter 6.
Here is another way of establishing the equality: Assume for the moment
that K(0) is ¬nite, and that, as we have already noted, f (0) = 0. Then,
d x
K(x ’ y)f (y) dy (9.43)
dx 0
9.4. SEPARABLE KERNELS 263
is equal to
x
‚x K(x ’ y)f (y) dy,
K(0)f (x) +
0
x
= K(0)f (x) ’ ‚y K(x ’ y)f (y) dy
0
x x
= K(0)f (x) ’ ‚y K(x ’ y)f (y) dy + K(x ’ y)f (y) dy
0 0
x
= K(0)f (x) ’ K(0)f (x) ’ K(x)f (0) + K(x ’ y)f (y) dy
0
x
K(x ’ y)f (y) dy.
= (9.44)
0
Since K(0) cancelled out, we need not worry that it is divergent! More
rigorously, we should regularize the improper integral by raising the lower
limit on the integral to a small positive quantity, and then taking the limit
that this goes to zero at the end of the calculation.
9.4 Separable Kernels
Let
N
K(x, y) = pi (x)qi (y), (9.45)
i=1
where {pi } and {qi } are two linearly independent sets of functions. The
range of K is therefore the span pi of the set {pi }. Such kernels are said
to be separable. The theory of integral equations containing such kernels is
especially transparant.
9.4.1 Eigenvalue problem
Consider the eigenvalue problem
»u(x) = K(x, y)u(y) dy (9.46)
D
for a separable kernel. Here D is some range of integration, and x ∈ D If
» = 0, we know that u has to be in the range of K, so we can write
u(x) = ξi pi (x). (9.47)
i
264 CHAPTER 9. INTEGRAL EQUATIONS
Inserting this into the integral, we ¬nd that our problem reduces to the ¬nite
matrix eigenvalue equation
»ξi = Aij ξj , (9.48)
where
Aij = qi (y)pj (y) dy. (9.49)
D
Matters are especially simple when qi = p— . In this case Aij = A— so the
i ji
matrix A is Hermitian, and therefore has N linearly independent eigenvec
tors. Observe that none of the N associated eigenvalues can be zero. To see
this, suppose that v(x) = i ζi pi (x) is an eigenvector with zero eigenvalue.
In other words, suppose that
p— (y)pj (y)ζj dy.
0= pi (x) (9.50)
i
D
i
Since the pi (x) are linearly independent, we must have
p— (y)pj (y)ζj dy = 0,
0= (9.51)
i
D
for each i separately. Multiplying by ζi— and summing we ¬nd
pj (y)ζj 2 dy,

0= (9.52)
D j
and v(x) itself must have been zero. The remaining (in¬nite in number)
eigenfunctions span qi ⊥ and have » = 0.
9.4.2 Inhomogeneous problem
It is easiest to discuss inhomogeneous separablekernel problems by example.
Consider the equation
1
u(x) = f (x) + µ K(x, y)u(y) dy, (9.53)
0
where K(x, y) = xy. Here, f (x) and µ are given, and u(x) is to be found.
We know that u(x) must be of the form
u(x) = f (x) + ax, (9.54)
9.4. SEPARABLE KERNELS 265
and the only task is to ¬nd the constant a. We plug u into the integral
equation and, after cancelling a common factor of x, we ¬nd
1 1 1
y 2 dy.
a=µ yu(y) dy = µ yf (y) dy + aµ (9.55)
0 0 0
The last integral is equal to µa/3, so
1 1
a 1’ µ =µ yf (y) dy, (9.56)
3 0
and ¬nally
µ 1
u(x) = f (x) + x yf (y) dy. (9.57)
(1 ’ µ/3) 0
Notice that this solution is meaningless if µ = 3. We can relate this to the
eigenvalues of the kernel K(x, y) = xy. The eigenvalue problem for this
kernel is
1