34032_IntEqunsNeumannSeries

MATH34032: Green’s Functions, Integral Equations and the Calculus of Variations 1

Section 4

Integral Equations: Neumann Series

As we have seen in Section 3, a Fredholm integral equation of the second kind may bewritten as

f (x) = λ

∫ b

a

K (x, y) f (y)dy + g (x) . (4.1)

We also saw, in section 3, that if the Kernel is separable, then we may take steps to solvethe integral equation analytically, by reducing it to a system of linear equations. However,what if the Kernel is not separable? In general these problems are much harder. Oneapproach exploits the fact that for sufficiently small values of λ it looks like we can neglectthe integral term to obtain the approximate solution

f (x) ≃ g (x) = f0 (x) . (4.2)

We can then improve on this approximation by substituting (4.2) into the right hand sideof (4.1), i.e.

f (x) ≃ f0 (x) + λ

∫ b

a

K (x, y) f0 (y) dy

= f0 (x) + λf1 (x) . (4.3)

Substituting (4.3) into the right hand side of (4.1) then yields an even better approximation

f (x) ≃ f0 (x) + λ

∫ b

a

K (x, y) [f0 (y) + λf1 (y)] dy

= f0 (x) + λ

∫ b

a

K (x, y) f0 (y)dy + λ2

∫ b

a

K (x, y) f1 (y) dy

= f0 (x) + λf1 (x) + λ2

∫ b

a

K (x, y) f1 (y) dy

= f0 (x) + λf1 (x) + λ2f2 (x) ,

say. This suggests letting

f (x) =∞

∑

n=0

λnfn (x) , (4.4)

which may be substituted into (4.1),

∞∑

n=0

λnfn (x) = λ

∫ b

a

K (x, y)

[

∞∑

n=0

λnfn (y)

]

dy + g (x)

=∞

∑

n=0

λn+1

∫ b

a

K (x, y) fn (y) dy + g (x) .

Equating coefficients of like powers in λ gives

λ0 : f0 (x) = g(x),

λn, n > 0 : fn (x) =∫ b

aK (x, y) fn−1 (y) dy.

(4.5)


Definition 4.1 Formula (4.4), with (4.5), is called the Neumann series for the integralequation (4.1).

Example 1: Find the first 3 terms of the Neumann series for the integral equation

u(x) = 1 + ǫ

∫

1

0

(x − y)u(y) dy.

where ǫ ∈ R.

We note that λ = ǫ and K(x, y) = (x − y) so that

u(x) = u0(x) + ǫu1(x) + ǫ2u2(x) + ....

and it remains for us to determine each term. Clearly neglecting the integral term givesu0(x) = 1, and then

u1(x) =

∫

1/2

0

(x − y)u0(y)dy (4.6)

=

[

xy −y2

2

]1/2

0

(4.7)

=

(

x

2−

1

8

)

(4.8)

and then

u2(x) =

∫

1/2

0

(x − y)u1(y)dy (4.9)

=

∫

1/2

0

(x − y)

(

y

2−

1

8

)

dy (4.10)

=

[

xy2

4− x

y

8−

y3

6+

y2

16

]1/2

0

(4.11)

= −1

192(4.12)

so that the Neumann series takes the form

u(x) = 1 + ǫ

(

x

2−

1

8

)

−ǫ2

192+ .... (4.13)

But this approach is clearly only useful if the resulting series converges.....


Iterated kernels and convergence

If the Neumann series is convergent it solves (4.1) uniquely. Now

fn (x) =

∫ b

a

K (x, y1) fn−1 (y1) dy1

=

∫ b

a

∫ b

a

K (x, y1) K (y1, y2) fn−2 (y2) dy2dy1,

replacing y by y1 in the first integral, and y2 in the second integral (for fn−1). We cancontinue replacing fm(x) by an integral containing fm−1(x) for each m down to m = 1,i.e.

fn(x) =

∫ b

a

∫ b

a

. . .

∫ b

a

K (x, y1)K (y1, y2) . . .K (yn−1, yn) f0 (yn) dyn . . . dy2dy1

=

∫ b

a

{∫ b

a

. . .

∫ b

a

K (x, y1) K (y1, y2) . . .K (yn−1, yn) dyn−1 . . . dy1

}

f0 (yn) dyn

=

∫ b

a

Kn (x, y) g (y) dy. (4.14)

Definition 4.2 The nth iterated kernel is

Kn (x, y) =

∫ b

a

. . .

∫ b

a

K (x, y1)K (y1, y2) . . .K (yn−1, y)dyn−1 . . . dy1.

Proposition 4.3 The nth iterated kernel satisfies the following inductive definition

K1 (x, y) = K (x, y) ,

Kn+1 (x, y) =∫ b

aK (x, z) Kn (z, y) dz.

Proof: Using Definition 4.2 with n replaced by n + 1 we have

Kn+1 (x, y) =

∫ b

a

. . .

∫ b

a

K (x, y1) K (y1, y2) . . .K (yn−1, yn) K (yn, y)dyn . . . dy1.

Now, setting y1 = z, say, and then relabelling each ym, m = 2, 3, . . . n, as ym−1 yields

Kn+1 (x, y) =

∫ b

a

K (x, z)

∫ b

a

. . .

∫ b

a

K (z, y1) . . .K (yn−2, yn−1) K (yn−1, y)dyn−1 . . . dy1dz,

or

Kn+1 (x, y) =

∫ b

a

K (x, z) Kn (z, y) dz

as required.From (4.4), (4.14) the Neumann series may be written

f (x) = f0 (x) +

∞∑

n=1

[

λn

∫ b

a

Kn (x, y) g (y)dy

]


= f0 (x) + λ

∫ b

a

[

∞∑

n=1

λn−1Kn (x, y)

]

g (y) dy

= g (x) + λ

∫ b

a

R (λ, x, y) g (y)dy. (4.15)

Thus, comparing with our previous definition of the resolvent kernel (see Definition 3.11)we get the following proposition.

Proposition 4.4 The resolvent kernel R (λ, x, y) may be written

R (λ, x, y) =

∞∑

n=1

λn−1Kn (x, y) .

Theorem 4.5: If it exists, R (λ, x, z) is unique and solves (4.1) via (4.15).

No proof given.

Theorem 4.6: The Neumann series for the Fredholm equation

f (x) = λ

∫ b

a

K (x, y) f (y)dy + g (x) ,

where K (x, y) and g (x) are bounded and absolutely integrable, converges absolutely for allvalues of λ such that

|λ| <1

M (b − a),

whereM = sup

a≤x≤ba≤y≤b

|K (x, y)| .

No proof given.

Example 2: Determine for which values of ǫ the Neumann series converges in Example1.

Using Theorem 4.6, we see that b = 1, a = 0 and

M = supa≤x≤ba≤y≤b

|K (x, y)|

= supa≤x≤ba≤y≤b

|x − y|

= 1

so that the series converges for |ǫ| < 1.


Theorem 4.7: The Neumann series for the Volterra equation

f (x) = λ

∫ x

a

K (x, y) f (y) dy + g (x) ,

where K (x, y) is bounded and g (x) is absolutely integrable, is absolutely convergent for allvalues of λ.

No proof given.

Theorem 4.8: The mth iterated kernel for a Volterra integral equation is given by

Km (x, y) =

{∫ x

yK (x, z) Km−1 (z, y) dz if y < x,

0 if y ≥ x.

Proof: We know from the definition of Volterra kernels that K1(x, y) = K(x, y) is zerofor y > x. So, start with m = 2:

K2 (x, y) =

∫ b

a

K (x, z) K (z, y) dz.

The first term in the integrand, K(x, z), is zero when z > x and the second, K(z, y), iszero when z < y. Hence

K2 (x, y) =

∫ x

y

K (x, z) K1 (z, y) dz.

The proof may now be completed by induction.

Example 3: Solve the Volterra integral equation

f (x) = λ

∫ x

0

f (y) dy + g (x)

and write down the resolvent kernel.

Solution: Here, K1 (x, y) = K (x, y) = 1.

K2 (x, y) =

∫ x

y

K (x, z) K1 (z, y) dz =

∫ x

y

dz = x − y,

K3 (x, y) =

∫ x

y

K (x, z) K2 (z, y) dz =

∫ x

y

1. (z − y)dz

=

[

1

2(z − y)2

]x

y

=1

2(x − y)2

,

K4 (x, y) =

∫ x

y

K (x, z) K3 (z, y) dz =

∫ x

y

1.1

2(z − y)2 dz

=

[

1

3!(z − y)3

]x

y

=1

3!(x − y)3 .


Suppose that, in general,

Kn (x, y) =(x − y)n−1

(n − 1)!. (4.16)

Assume true for n = k, then for n = k + 1

Kk+1 =

∫ x

y

K (x, z) Kk (z, y) dz =

∫ x

y

(z − y)k−1

(k − 1)!dz

=

[

(z − y)k

k!

]x

y

=(x − y)k

k!.

Hence, (4.16) follows by induction. Therefore, from (4.15)

f (x) = g (x) +

∞∑

n=1

[

λn

∫ x

0

Kn (x, y) g (y) dy

]

= g (x) + λ

[

∫ x

0

∞∑

n=1

λn−1Kn (x, y) g (y)dy

]

= g (x) + λ

∫ x

0

[

∞∑

n=1

λn−1(x − y)n−1

(n − 1)!

]

g (y) dy

= g (x) + λ

∫ x

0

exp [λ (x − y)] g (y)dy.

Note, by comparison with (4.15), that the resolvent kernel is exp [λ (x − y)].The theory for degenerate kernels and for Neumann series may be used to prove thefollowing theorem, valid for an arbitrary kernel.

Theorem 4.9: The Fredholm Alternative for integral equations

Either the integral equation

f − λK̂f = g, f (x) − λ

∫ b

a

K (x, y) f (y)dy = g (x) , (4.17)

for fixed λ has precisely one solution f (x) for each arbitrary g (x); in particularf = 0 when g = 0.

The associated adjoint integral equation

f (x) − λ

∫ b

a

K (y, x) f (y) dy = g (x) ,

also has a unique solution. Note that the adjoint equation has kernel K(y, x) notK(x, y).

In this case λ is called a regular value.


Or the associated homogeneous equation

f = λK̂f, f (x) = λ

∫ b

a

K (x, y) f (y)dy,

has m > 0, m ∈ N, linearly independent solutions f1, f2, . . . , fm.

The associated homogeneous adjoint integral equation

f (x) = λ

∫ b

a

K (y, x) f (y) dy,

also has m linearly independent solutions h1, h2, . . . , hm.

In this case equation (4.17) has a solution if and only if g (x) satisfies

〈g, hi〉 =

∫ b

a

g (y)hi (y)dy = 0

for i = 1, . . . , m, and the solution of (4.17) is then given by

f (x) = f0 (x) +

m∑

i=1

αifi (x)

where the αi are arbitrary constants.

In this case λ is called a characteristic value or eigenvalue.

34032_IntEqunsNeumannSeries

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