MATH 551 LECTURE NOTES FREDHOLM (A BRIEF INTRODUCTION)

Topics covered • Fredholm integral operators ◦ Integral equations (Volterra vs. Fredholm) ◦ Eigenfunctions for separable kernels ◦ Adjoint operator, symmetric kernels • Solution procedure (separable) ◦ Solution via eigenfunctions (first and second kind) ◦ Shortcuts: undetermined coefficients ◦ An example (separable kernel, n = 2) • Non-separable kernels (briefly) ◦ Hilbert-Schmidt theory

Preface Read the notes before proceeding. This is covered in the book (Section 9.4), but the material on integral equations is not. For references on integral equa- tions (and other topics covered in the book too!), see: • Riley and Hobson, Mathematical methods for physics and engineering (this is an extensive reference, also for other topics in the course) • Guenther and Lee, Partial differential equations of mathematical physics and integral equations (more technical; not the best first reference) • J.D. Logan, Applied (more generally about applied mathematics tech- niques, with a good section on integral equations)

1. Fredholm integral equations: introduction Differential equations Lu = f are a subset of more general equations involving linear op- erators L. Here, we give a brief treatment of a generalization to integral equations.

To motivate this, every ODE IVP can be written as an ‘integral ’ by integrating. For instance, consider the first order IVP du = f(x, u(x)), u(a) = u . (1.1) dx 0 Integrate both sides from a to x to get the integral equation Z x u(x) = u0 + f(s, u(s)) ds. (1.2) a If u solves (1.2) then it also solves (1.1); they are ‘equivalent’ in this sense. However, it is slightly more general as u does not need to be differentiable. 1 2 INTEGRAL EQUATIONS I

Definition: A Volterra integral equation for u(x) has the form Z x u(x) = a + k(x, t)u(t) dt. a The k(x, t) is the kernel. Note the integral upper bound is x (the ind. variable).

Volterra integral equations are ‘equivalent’ to ODE initial value problems on x ≥ a for linear ODEs. They will not be studied here.

In contrast, ODE boundary value problems generalize to Fredholm integral equations. Such an equation involves an integral over the whole domain (not up to x):

Definition: A Fredholm integral equation (FIE) has two forms: • First kind: (FIE-1) Z b k(x, t)u(t) dt = f(x) (1.3) a • Second kind: (FIE-2) Z b γu(x) + k(x, t)u(t) dt = f(x) (1.4) a The ‘second kind’ is the first kind plus a multiple of u. Note that the operator Z b Lu = γu + k(x, t)u(t) dt a is linear; this is called a Fredholm integral operator.

The simplest kernels are separable (‘degenerate’), which have the form n X k(x, t) = αj(x)βj(t) = finite sum of ‘separated’ products. (1.5) j=1 More complicated kernels are non-separable. Examples include: • k = 1/(x − t) (Hilbert transform) • k = e−ixt (Fourier transform) • k = e−xt () We will study the last two later. Writing a non-separable kernel in the form (1.5) would require an infinite series, which complicates the analysis. Instead, we will utilize an integral form and some complex analysis later to properly handle non-separable kernels. INTEGRAL EQUATIONS I 3

2. Solving separable FIEs Suppose k is separable and L is a Fredholm integral operator of the first kind: n Z b X Lu = k(x, t)u(t) dt, k(x, t) = αj(x)βj(t). (2.1) a j=1 We wish to solve the eigenvalue problem Lφ = λφ. This can always be done for a separable kernel. The main result is the following:

Eigenfunctions for FIE-1: For the FIE (2.1) with separable kernel,

1) there are n non-zero eigenvalues λ1, ··· , λn with eigenfunctions φ1, ··· , φn. Each P eigenfunction is a linear combination of the αj’s, i.e. φj(x) = cjkαk(x) 2) λ∞ = 0 is an eigenvalue with infinite multiplicity: an infinite set of (orthogonal) ∞ eigenfunctions φm (m = 1, 2, ··· ) characterized by Z b ∞ ∞ 0 = hφm , βji = φm (t)βj(t) dt for j = 1, 2, ··· n. (2.2) a The eigenfunctions from (1) and (2) together are a basis for L2[a, b].

Proof, part 1 (non-zero eigenvalues): Consider λ 6= 0; look for an eigenfunction n X φ = cjαj. j=1 R b Derivation: Plug in this expression into the operator Lφ = a k(x, t)φ(t) dt: n ! n Z b X X Lφ = αi(x)βi(t) cjαj(t) dt a i=1 j=1 X Z b  = cjαi(x) βi(t)αj(t)dt 1≤i,j≤n a X = Aijcjαi(x) i,j R b where Aij = a βi(t)αj(t) dt. Now set this equal to n X λφ = ciαi(x). i=1

The coefficients on αi must match, leaving the linear system Z b T Ac = λc, c = (c1, ··· , cn) ,Aij = βi(t)αj(t) dt. (2.3) a so the eigenvalue problem for L is equivalent to a matrix eigenvalue problem for an n×n matrix. Assuming A is invertible, the result follows. 4 INTEGRAL EQUATIONS I

Claim 2 (zero eigenvalue): We show that λ = 0 is always an eigenvalue with infinite multiplicity. Observe that n n X Z b  X Lφ = αj(x) βj(t)φ(t) dt = hβj, φiαj(x). j=1 a j=1 This must be true for all x, so it follows that

Lφ = 0 ⇐⇒ hβj, φi = 0 for j = 1, ··· , n. That is, the set of eigenfunctions for λ = 0 is precisely the space of functions orthogonal to 2 the span of the βj’s. But L [a, b] is infinite dimensional and the span of {βj} has dimension n, so the orthogonal complement is infinite dimensional - this proves the claim. That is, 2 L [a, b] = span({βj}) + {Lφ = 0}. The fact that the basis for {Lφ = 0} is countable (a sequence indexed by n) also follows 2 1 ∞ from the fact that L [a, b] has this property. The eigenfunctions φm can be constructed using Gram-Schmidt (see example in subsection 2.1). 2.1. Example (eigenfunctions). Define the (separable) kernel k(x, t) = 4xt − 5x2t2, 2 2 (2.4) (α1 = x, α2 =x , β1 = 4t, β2 = −5t ) and consider the FIE of the first kind Z 1 Lu := k(x, t)u(t) dt = f. 0 The result says L has n = 2 non-zero eigenvalues. Look for an eigenfunction

φ = c1α1 + c2α2 (2.5) then plug in and integrate to get Z 1 (α1(x)β1(t) + α2(x)β2(t)) (c1α1(t) + c2α2(t)) dt = λ(c1α1(x) + c2α2(x)) 0

which gives (equating coefficients of α1(x), α2(x) on the LHS/RHS as in (2.3)) R 1 R 1      0 β1α1 dt 0 β1α2 dt c1 c1 1 1 = λ . R R c2 c2 0 β2α1 dt 0 β2α2 dt Plugging in the values from (2.4), the eigenvalue problem is  4/3 1  Ac = λc,A = (2.6) −5/4 −1 =⇒ λ = 1/2, c = (−6, 5)T λ = −1/6, c = (2, −3)T . Translating back to the integral equation with φ given by (2.5), the non-zero λ’s and φ’s are 2 2 {λ1 = 1/2, φ1 = −6x + 5x }, {λ2 = −1/6, φ2 = 2x − 3x } ∞ ∞ ∞ along with the zero eigenvalue and eigenfunctions {λ = 0, φ1 , φ2 , ···}.

1In terms, it is a ‘separable ’. INTEGRAL EQUATIONS I 5

Zero eigenvalue (details): For λ = 0, use the characterization (2.2): Z 1 Z 1 2 Lφ = 0 ⇐⇒ hφ, β1i = hφ, β2i = 0 ⇐⇒ tφ(t) dt = t φ(t) dt = 0. 0 0

First, we need to orthogonalize. Define p1 = t and ht2, ti 3 p = t2 − t = t2 − t. 2 ht, ti 4 It follows that if g(t) is any function then a zero eigenfunction is 2 X hg, pji φ∞ = g − p . hp , p i j j=1 j j 3 The whole set can be found using Gram-Schmidt, e.g. with p3 = 1, p4 = t , ··· . For instance, 2 X h1, pji 3 −1/24 3 10 φ = 1 − p = 1 − t − (t2 − t) = 1 − 4t + t2 hp , p i j 2 1/80 4 3 j=1 j j is an eigenfunction for λ = 0 (orthogonal to t and t2).

2.2. Adjoint operator. The operator L is not self-adjoint in general. To solve FIEs with ∗ an eigenfunction expansion, we need the adjoint L and its eigenfunctions ψn. To obtain it, 2 R b let h·, ·i denote the L inner product hf, gi = a f(x)g(x) dx.

We look for an integral operator L∗ such that

hLu, vi = hu, L∗vi for all u, v ∈ L2[a, b]. (2.7)

To find the adjoint, carefully exchange integration order (pay attention to which of t or x is the ‘integration variable’ here!). Compute

Z b Z b  hLu, vi = k(x, t)u(t) dt v(x) dx a a Z b Z b = k(x, t)u(t)v(x) dt dx a a Z b Z b  = k(x, t)v(x) dx u(t) dt a a = hu, L∗vi

Note the inner integration variable above is now x and not t. From this result, we see that

Z b Z b (L∗v)(t) = k(x, t)v(x) dx =⇒ (L∗v)(x) = k(t, x)v(t) dt a a R b after swapping symbols for t and x. Comparing to Lu = a k(x, t)u(t) dt, we have the following characterization for the adjoint: 6 INTEGRAL EQUATIONS I

FI operator adjoint: The FI operator of the first kind and its adjoint are Z b Z b Lu = k(x, t)u(t) dt, L∗v = k(t, x)v(t) dt (2.8) a a i.e. the adjoint has kernel k(t, x) (the kernel of L with the variables swapped). Moreover, L = L∗ ⇐⇒ k(x, t) = k(t, x). That is, L is self-adjoint if and only if the kernel is symmetric (e.g. k(t, x) = et+x).

3. Solving FIEs of the first kind (separable) We are now equipped to solve Lu = f (3.1) where L has a separable kernel (1.5). Note that from (2.8), it follows that L∗ also has a separable kernel.

∗ Thus the adjoint L also has n non-zero eigenvalues with eigenfunctions ψ1, ··· , ψn (same structure). Moreover, the ψ’s and φ’s are bi-orthogonal. That is, hφj, ψki = 0 for j 6= k ∞ ∞ ∞ and the same for the φm ’s and ψm ’s and between the two sets (e.g. hφj, ψm i = 0 for all j, m).

2 Reminder: to get the coefficient of φj, take the L inner product with ψj, e.g. X f = cjφj =⇒ hf, ψji = cjhφj, ψji. j

Solving the FIE: Project by taking h·, ψji to get

hLu, ψji = hf, ψji. We need to write Lu and f in terms of the eigenfunctions.

• For the RHS (f): This is direct:

X X ∞ ∞ f = fjφj + fm φm . (3.2) j m

Let kj = hφj, ψji. Take the inner product with ψj to get

fj = hf, ψji/kj. The other coefficients will only be needed later.

• For the LHS (Lu): First write

X X ∞ ∞ u = cjφj + cm φm . (3.3) j m The second term is sent to zero by L: X X Lu = cjLφj = λjcjφj. j j INTEGRAL EQUATIONS I 7

Taking the inner product with ψj, we get

hLu, ψji = λjcjhφj, ψji. • Combining: Equating the two parts, we find that

cj = fj = hf, ψji/kj, j = 1, ··· , n. ∞ ∗ Solvability: What about the cm ’s? The adjoint L has a zero eigenvalue, so we are in FAT case (B). A solvability condition decides between no solution or infinitely many solutions.

To find this condition, project onto one a eigenfunction for λ = 0, i.e. take the inner ∞ product of Lu = f with one of the ψm ’s: ∞ ∞ ∗ ∞ hf, ψm i = hLu, ψm i = hu, L ψm i = 0. ∞ Thus a solution exists only when f(x) has no φm component for any m, i.e. if and only if n X f = fjφj. (3.4) j=1

Summary of solution (separable FIE-1): If f is of the form (3.4), the solution is n X X hf, ψji u = c φ + c∞φ∞, c = j j m m j hφ , ψ i j=1 m j j ∞ and the cm ’s are arbitrary. If f is not of the form (3.4) then there is no solution.

3.1. The shortcut. There is a way to get around the full expansion. Recall that X φj(x) = d`,jα`(x) ` i.e each eigenfunction (for non-zero λ) is a linear combination of the α’s. Then ! X X X X Lu = λjcjφj = λjcj d`,jα`(x) = C`α`(x). j j ` ` It follows that we could anticipate the solution and guess X u = Cjαj j knowing that the result for u will be in this form. We lose the bi-orthogonal structure (no eigenfunctions), but the φ’s are not needed to calculate the solution. If n is small, the shortcut is much easier (the linear system is small). 3.2. Example: Return to the previous example (subsection 2.1), Z 1 Lu := k(x, t)u(t) dt = f, 0 with separable kernel 2 2 2 2 k(x, t) = 4xt − 5x t , (α1 = x, α2 = x , β1 = 4t, β2 = −5t ), 8 INTEGRAL EQUATIONS I

and non-zero eigenvalues/functions 2 2 {λ1 = 1/2, φ1 = −6x + 5x }, {λ2 = −1/6, φ2 = 2x − 3x }. ∞ along with the zero eigenvalue and eigenfunctions {φm }. Suppose the problem is Lu = 8x − 7x2. Note that λ = 0 is always an eigenvalue, so we are in case (B) of the FAT. Here, the RHS is 2 in the span of φ1 and φ2 (both linear combinations of x and x ), so there is a solution. It is P ∞ ∞ unique up to an arbitrary sum of the form m cm φm .

Direct method: Note that the kernel is symmetric here. Thus L is self-adjoint and ψj = φj. Write the solution as X ∞ ∞ u = a1φ1 + a2φ2 + cm φm . m Take the inner product of the equation with φj:

∗ hf, φji hf, φji = hLu, φji = hu, L φji = λjajhφj, φji =⇒ aj = . hφj, φji The coefficients can be computed from here; then X ∞ ∞ u = a1φ1 + a2φ2 + cm φm m ∞ for arbitrary cm ’s (details not carried out here).

Shortcut method: Instead look for the ‘unique part’ as a linear combination of α’s:

u = b1α1(x) + b2α2(x). Plug into the equation and write as a linear system: R 1 R 1            0 β1α1 dt 0 β1α2 dt b1 8 4/3 1 b1 8 1 1 = =⇒ = R R b2 −7 −5/4 −1 b2 −7 0 β2α1 dt 0 β2α2 dt ∞ so b1 = 12 and b2 = −8. The solution is then (for arbitrary cm ’s) 2 X ∞ ∞ u(x) = 12x − 8x + cm φm . m INTEGRAL EQUATIONS I 9

4. FIEs of the second kind (still separable) Now consider the FI operator of the second kind,

Lu = L1u + γu (4.1)

where L1 is an FIE of the first kind with a separable kernel: n Z b X L1u = k(x, t)u(t) dt, k(x, t) = αj(x)βj(t). a j=1

Note that the γu term added to L1u only shifted the eigenvalues. The result:

For the second kind: For the FIE of the second kind (4.1), the eigenfunctions are the same as L1 and the eigenvalues are shifted by γ. That is, the eigenvalues of L are ∞ λ1, ··· , λn (6= γ), λ = γ ∞ with the same eigenfunctions φ1, ··· , φn and φm (m = 1, 2, ··· ).

It follows that L has all non-zero eigenvalues unless −γ is an eigenvalue of L1.

Now we solve the FIE Lu = γu + L1u = f (4.2) under the assumption that λ1, ··· , λn 6= 0. (4.3) If (4.3) holds then the FAT case (A) applies (since λ∞ is non-zero) and there is a unique solution, unlike an FIE of the first kind.

Direct method (not ideal): Expand u and f in terms of the eigenfunctions: n n X X ∞ ∞ X X ∞ ∞ u = cjφj + cm φm , f = fjφj + fm φm . j=1 m j=1 m Now plug into the FIE (4.2) to get n ! n X X ∞ ∞ X ∞ X ∞ ∞ Lu = L cjφj + cm φm = λjcjφj + λ cm φm . j=1 m j=1 m Equate this to f (note that now λ∞ = γ 6= 0) to get the coefficients: n X ∞ X ∞ ∞ λjcjφj + λ cm φm = f =⇒ · · · j=1 m This can be continued, but there is a better shortcut, inspired by the FIE-1 case. The trick P is that the ‘infinite part’ (with m(··· )) of the solution is really a multiple of f, leaving only the ‘finite part’, which is a sum only over n terms.

Practical note: The direct method leads to trivial equations (from orthogonality), but P there are an infinite number of them. Moreover, the m part is best avoided if we can get away with a solution containing only a finite sum! 10 INTEGRAL EQUATIONS I

4.1. Shortcut (undetermined coefficients): Assume a solution of the form n X 1 u = b α (x) + f(x). (4.4) j j p j=1

Since the kernel is separable, L1g for any function g(t) is a linear combination of αj(x)’s: Z b Z b X X Z b  L1g = k(x, t)g(t) dt = αj(x)βj(t)g(t) dt = βj(t)g(t) dt αj(x). (4.5) a a j j a Now plug the guess (4.4) into Lu and keep the f part of γu separate from the rest. All the other terms, via rule (4.5), will give various linear combinations of αj’s. To be precise, X X L1αi = cijαj(x),L1f = djαj(x). j j

Now plug in and collect all the αj terms and the leftover f part:

Lu = γu + L1u X γ X 1 = γb α (x) + f + b (L α ) + L f j j p i 1 i p 1 j i ! X γ X X = γb α (x) + f + c b + d α (x) j j p ij j j j j j i ! X X X γ = γb α (x) + c b + d α (x) + f j j ij j j j p j j i |{z} | {z } f =0 P That is, the (··· )αj part must cancel, leaving just f, so p = γ, X bj = cijbj + dj for j = 1, 2, ··· , n i which is linear system for b = (b1, ··· , bn) of the form b = Cb + d. Now recall that under the assumptions made at the start, the FAT guarantees a unique solution, so we are done (the guess is a solution, so it is the unique solution).

4.2. Example: again, return to the example. Now let Z 1 1 L1u = k(x, t)u(t) dt, Lu = u + L1u. 0 2

The eigenvalues of L are 1/2 + 1/2 and 1/2 − 1/6, with the same eigenfunctions as L1 computer earlier. Consider 1 u + L u = ex. 2 1 Look for a solution 1 u = b α (x) + b α (x) + ex. 1 1 2 2 p INTEGRAL EQUATIONS I 11

After following the process2 (plug everything in), p = 2 and u = (72 − 30e)x + (55e − 140)x2 + 2ex. This is the unique solution! Note the 2ex deals neatly with the RHS; no infinite sum required. The downside is that the algebra for b1 and b2 is messy, but it is a small linear system, and a computer algebra package has no trouble with it.

5. What if the kernel is non-separable? For a non-separable kernel such as ∞ X k(x, t) = αj(x)βj(t) (5.1) j=1 the theory is different. More conditions are required to ensure that the eigenvalue problem behaves well. If the kernel is ‘nice’, there is an infinite set of eigenvalues with structure similar to Sturm-Liouville theory. As for Sturm-Liouville operators the general idea3 is that L self-adjoint + (L) is ‘nice’ =⇒ good properties for eigenfunctions/values. For integral equations, Hilbert-Schmidt theory provides the answer for what is required to be ‘nice’ and what ‘good properties’ result. The Hilbert-Schmidt condition is Z b Z b k(x, t)2 dx dt < ∞ (5.2) a a R b which guarantees that the operator a k(x, t)f(t) dt is ‘nice’ (precisely, a compact operator on L2[a, b]). Hilbert-Schmidt theory then says that if Z b Lu = u(t)k(x, t) dt a satisfies the conditions 1) L is self-adjoint (in L2[a, b]) 2) k(x, t) is not separable (although the separable case is similar) 3) The Hilbert-Schmidt condition (5.2) holds ( =⇒ L is ‘compact’) then the eigenvalue problem has nice properties: • The eigenvalues λ are real • The eigenvalues are a discrete sequence, each with finite multiplicity:

|λ1| ≤ |λ2| ≤ · · · 2 P 2 • The eigenfunctions {φj} are an orthogonal basis for L [a, b](f = cjφj if f ∈ L [a, b]) • The Fredholm Alternative and eigenfunction expansions for Lu = f apply (in the same way that we used them for BVPs with Sturm-Liouville theory) For further reading, see the references at the start of the notes. We will study non-separable kernels in a different way when addressing the Fourier and Laplace transforms.

2The computations are messy and not included here. 3Both the theory for integral operators and Sturm-Liouville theory are part of a more general theory; the structure comes from the theory of compact operators on a Hilbert space (the ‘spectral theorem for compact operators’).