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Math 551 Lecture Notes Fredholm Integral Equations (A Brief Introduction)

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Math 551 Lecture Notes Fredholm Integral Equations (A Brief Introduction) MATH 551 LECTURE NOTES FREDHOLM INTEGRAL EQUATIONS (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 Fredholm alternative 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 mathematics (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 equation' 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 function 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 (Laplace transform) 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) λ1 = 0 is an eigenvalue with infinite multiplicity: an infinite set of (orthogonal) 1 eigenfunctions φm (m = 1; 2; ··· ) characterized by Z b 1 1 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 fβjg has dimension n, so the orthogonal complement is infinite dimensional - this proves the claim. That is, 2 L [a; b] = span(fβjg) + fLφ = 0g: The fact that the basis for fLφ = 0g is countable (a sequence indexed by n) also follows 2 1 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 fλ1 = 1=2; φ1 = −6x + 5x g; fλ2 = −1=6; φ2 = 2x − 3x g 1 1 1 along with the zero eigenvalue and eigenfunctions fλ = 0; φ1 ; φ2 ; · · · g: 1In functional analysis terms, it is a `separable Hilbert space'. 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 φ1 = 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 n. 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 2 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.
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