Integral equations

In this lecture we will study the Fredholm equation defined as follows. Let the integral operator with a kernel K(x,y) defined on [a,b]×[a,b] be defined as before:

for a function f on [a,b]. A special case is given by Volterra equation by an operator integral operator (54) T with a kernel K(x,y)=0 for all y>x which could be written as:

We will consider integral operators with kernels K such that ∫_a^b∫_a^b K(x,y) d x d y<∞, then by Theorem 15 T is a Hilbert–Schmidt operator and in particular bounded.

As a reason to study Fredholm operators we will mention that solutions of differential equations in mathematical physics (notably heat and wave equations) requires a decomposition of a function f as a linear combination of functions K(x,y) with “coefficients” φ. This is an continuous analog of a discrete decomposition into Fourier series.

Using ideas from the proof of Lemma 4 we define Neumann series for the resolvent:

Example 1 Solve the Volterra equation

φ(x)−λ

∫

y φ(y) d y=x², on L₂[0,1].

In this case I−λ T φ = f, with f(x)=x² and:

K(x,y)=

⎧
⎨
⎩

y,	0≤ y ≤ x;
0,	x< y ≤ 1.

Straightforward calculations shows:

(Tf)(x)

∫

y· y² d y=

x⁴

(T²f)(x)

∫

y⁴

 d y=

x⁶

, …

and generally by induction:

(Tⁿf)(x) =

∫

y²ⁿ

2ⁿ⁻¹n!

 d y=

x²ⁿ⁺²

2ⁿ(n+1)!

Hence:

φ(x)

∞

∑

λⁿTⁿ f =

∞

∑

λⁿx²ⁿ⁺²

2ⁿ(n+1)!

∞

∑

λⁿ⁺¹x²ⁿ⁺²

2ⁿ⁺¹(n+1)!

(e^λ x²/2−1) for all λ ∈ ℂ∖ {0},

because in this case r(T)=0. For the Fredholm equations this is not always the case, see Tutorial problem 29.

Among other integral operators there is an important subclass with separable kernel, namely a kernel which has a form:

i.e. the image of T is spanned by g₁(x), …, g_n(x) and is finite dimensional, consequently the solution of such equation reduces to linear algebra.

Example 2 Solve the Fredholm equation (actually find eigenvectors of T):

φ(x)

2π

∫

cos(x+y)φ(y) d y

2π

∫

(cosxcosy − sinx siny)φ(y) d y.

Clearly φ (x) should be a linear combination φ(x)=Acos x+Bsinx with coefficients A and B satisfying to:

2π

∫

cosy (Acosy+Bsiny) d y,

−λ

2π

∫

siny (Acosy+Bsiny) d y.

Basic calculus implies A=λπ A and B=−λπ B and the only nonzero solutions are:

λ=π⁻¹	A ≠ 0	B = 0
λ=−π⁻¹	A = 0	B ≠ 0

Theorem 3 Suppose that K(x,y) is a continuous function on [a,b]×[a,b] and K(x,y)=K(y,x) and operator T is defined by (54). Then

T is a self-adjoint Hilbert–Schmidt operator.
All eigenvalues of T are real and satisfy ∑_n λ_n²<∞.
The eigenvectors v_n of T can be chosen as an orthonormal basis of L₂[a,b], are continuous for nonzero λ_n and
Tφ=
∞

∑

n=1

λ_n ⟨ φ,v_n ⟩v_n where φ=
∞

∑

n=1

⟨ φ,v_n ⟩v_n

Proof.

The condition K(x,y)=K(y,x) implies the Hermitian property of T:

⟨ Tφ,ψ ⟩

∫

⎛
⎜
⎜
⎝

∫

K(x,y)φ(y) d y

⎞
⎟
⎟
⎠

ψ(x) d x

∫

K(x,y)φ(y) ψ(x) d x d y

∫

φ(y)

⎛
⎜
⎜
⎝

∫

K(y,x) ψ(x)

 d x

⎞
⎟
⎟
⎠

⟨ φ,Tψ ⟩.

The Hilbert–Schmidt property (and hence compactness) was proved in Theorem 15.

Spectrum of T is real as for any Hermitian operator, see Theorem 2 and finiteness of ∑_n λ_n² follows from Hilbert–Schmidt property
The existence of orthonormal basis consisting from eigenvectors (v_n) of T was proved in Corollary 8. If λ_n≠ 0 then:
v_n(x₁)−v_n(x₂) = λ_n⁻¹((Tv_n)(x₁)−(Tv_n)(x₂))

=
1

λ_n

b

∫

a

(K(x₁,y)−K(x₂,y))v_n(y) d y

and by Cauchy–Schwarz-Bunyakovskii inequality:
⎪
⎪ v_n(x₁)−v_n(x₂) ⎪
⎪ ≤
1

⎪
⎪ λ_n ⎪
⎪

⎪⎪
⎪⎪ v_n ⎪⎪
⎪⎪ ₂
b

∫

a

⎪
⎪ K(x₁,y)−K(x₂,y) ⎪
⎪  d y

which tense to 0 due to (uniform) continuity of K(x,y).

□

Proof. Let φ=∑₁^∞a_n v_n where a_n=⟨ φ,v_n ⟩, then

φ−λ Tφ=

∞

∑

a_n(1−λ λ_n) v_n =f=

∞

∑

⟨ f,v_n ⟩v_n

if and only if a_n=⟨ f,v_n ⟩/(1−λ λ_n) for all n. Note 1−λ λ_n≠ 0 since λ⁻¹∉σ(T).

Because λ_n→ 0 we got ∑₁^∞| a_n |² by its comparison with ∑₁^∞| ⟨ f,v_n ⟩ |²=||f||², thus the solution exists and is unique by the Riesz–Fisher Theorem. □

Theorem 5 (Fredholm alternative) Let T∈K(H) be compact normal and λ∈ℂ∖ {0}. Consider the equations:

φ−λ Tφ	=	0	(60)
φ−λ Tφ	=	f	(61)

then either

the only solution to (60) is φ=0 and (61) has a unique solution for any f∈ H; or
there exists a nonzero solution to (60) and (61) can be solved if and only if f is orthogonal all solutions to (60).

Proof.

If φ=0 is the only solution of (60), then λ⁻¹ is not an eigenvalue of T and then by Lemma 6 is neither in spectrum of T. Thus I−λ T is invertible and the unique solution of (61) is given by φ=(I−λ T)⁻¹ f.
A nonzero solution to (60) means that λ⁻¹∈σ(T). Let (v_n) be an orthonormal basis of eigenvectors of T for eigenvalues (λ_n). By Lemma 2 only a finite number of λ_n is equal to λ⁻¹, say they are λ₁, …, λ_N, then
(I−λ T)φ=
∞

∑

n=1

(1−λ λ_n)⟨ φ,v_n ⟩v_n =
∞

∑

n=N+1

(1−λ λ_n)⟨ φ,v_n ⟩v_n.

If f=∑₁^∞⟨ f,v_n ⟩v_n then the identity (I−λ T)φ=f is only possible if ⟨ f,v_n ⟩=0 for 1≤ n≤ N. Conversely from that condition we could give a solution
φ=
∞

∑

n=N+1

⟨ f,v_n ⟩

1−λ λ_n

v_n +φ₀, for any φ₀∈Lin(v₁,…,v_N),

which is again in H because f∈ H and λ_n→ 0.

□

Example 6 Let us consider

(Tφ)(x)=

∫

(2xy−x−y+1)φ(y) d y.

Because the kernel of T is real and symmetric T=T^*, the kernel is also separable:

(Tφ)(x)=x

∫

(2y−1)φ(y) d y+

∫

(−y+1)φ(y) d y,

and T of the rank 2 with image of T spanned by 1 and x. By direct calculations:

↦

x +

or T is given by the matrix

⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠

According to linear algebra decomposition over eigenvectors is:

λ₁=

with vector

⎛
⎜
⎝

⎞
⎟
⎠

λ₂=

with vector

⎛
⎜
⎜
⎜
⎜
⎝

−

⎞
⎟
⎟
⎟
⎟
⎠

with normalisation v₁(y)=1, v₂(y)=√12(y−1/2) and we complete it to an orthonormal basis (v_n) of L₂[0,1]. Then

If λ≠ 2 or 6 then (I−λ T)φ = f has a unique solution (cf. equation (59)):

∑

n=1

⟨ f,v_n ⟩

1−λ λ_n

v_n +

∞

∑

n=3

⟨ f,v_n ⟩ v_n

∑

n=1

⟨ f,v_n ⟩

1−λ λ_n

v_n +

⎛
⎜
⎜
⎝

f−

∑

n=1

⟨ f,v_n ⟩ v_n)

⎞
⎟
⎟
⎠

∑

n=1

λλ_n

1−λ λ_n

⟨ f,v_n ⟩v_n.

If λ=2 then the solutions exist provided ⟨ f,v₁ ⟩=0 and are:
φ=f+
λλ₂

1−λ λ₂

⟨ f,v₂ ⟩v₂+Cv₁=f+
1

2

⟨ f,v₂ ⟩v₂+Cv₁, C∈ℂ.
If λ=6 then the solutions exist provided ⟨ f,v₂ ⟩=0 and are:
φ=f+
λλ₁

1−λ λ₁

⟨ f,v₁ ⟩v₁+Cv₂=f−
3

2

⟨ f,v₂ ⟩v₂+Cv₂, C∈ℂ.

10 Integral equations