XIV. Solving Y = KY + f.

Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

*(c) Copyright 1994,1997,2000 by Evans M. Harrell II and James V. Herod. All rights reserved.

version of 14 March 2000

Here is a Mathematica notebook with calculations for this chapter, and here are some Maple worksheets with similar calculations

  • Separable kernels
  • Small K(x,t)
  • Small K*(x,t)
  • Neither small nor separable

  • XIV. Solving Y = KY + f

    In this chapter we shall learn how to solve integral equations in three situations:

    These terms will be explained as they are encountered. First we discuss the separable case.

    Definition XIV.1 Suppose that there are an integer n and functions

    such that, for each p , ap and bp are in L2[0,1]. Then K has a separable kernel if its kernel is given by

    sum of a_p(x) b_p(t)                              (14.1)

    Another term for operators K of this type is finite-rank, and we shall see that they can be considered as matrices of finite rank.

    With the supposition that K is separable, it is not hard to find y such that y = Ky + f, for this equation can be re-written as

    y(x) = Sum a_p(x) integral(b_p(t)y(t) dt + f(x)                           (14.2)

    or, using the notation of inner products,

    y = sum of a_p(x) <b_p, y > + f(x)                           (14.3)

    We can see that if the sequence
    of functions on [0,1] is a linearly independent sequence, then y will have the following special form:

    there is a sequence {cp} of numbers such that

    Why is this? In (14.1), all the definite integrals over t are just numbers. Even though we do not know their values yet, we can call them cp and procede to determine their values with a bit of algebraic labor.


    y(x) = Sum c_p a_p(x)   +  f(x)

    Substitute this in the equation to be solved:


    and we see that

    c_p = sum <b_p,a_q> c_q + <b_p,f>

    This now reduces to a matrix problem:

    simul eqns for c_p

    Define K and f to be the matrix and vector so defined that the last equation is rewritten as

           c = K c + f.

    We now employ ideas from linear algebra. The equation c = K c + f has exactly one solution provided

    det( 1 - K ) != 0. The Fredholm alternative theorems, described in chapter XIII address these ideas. (Review the alternative theorems for matrices.) Once the sequence

    is found, we have a formula for y(x).

    Example XIV.2: In the exercises of chapter XIII, it should have been established that if

           K(x,t) = 1 + sin(pi x) cos(pi t),


           K*(x,t) = 1 + sin(pi t) cos(pi x).


            y = Ky has solution y(x) = 1


           y = K*y has solution y(x) = pi + 2 cos(pi x).

    It is the promise of the Fredholm Alternative theorems that

           y = Ky + f

    has a solution provided that

    Let us try to solve y = Ky + f and watch to see where the requirement that f should be perpendicular to the function pi +2 cos(pi t) appears.

    To solve y = Ky + f is to solve

    As usual we see that the solution must be of the form y(x) = a + b sin(pi x) + f(x), and substitute this for y:

    From this, we get the algebraic equations

    Hence, in our guess for y, we find that a can be anything and that b must be

    -(pi/2) int(f(t), t=0..1)

    and b must also be

    int(cos(pi t) f(t), t=0..1)

    The naive pupil might think this means there are two (possibily contradictory) requirements on b. The third of the Fredholm Alternative theorems assures the student that there is only one requirement!

    If the kernel is not separable, an alternative hypothesis that will enable one to solve the equation y = Ky + f is to suppose that the kernel for K is small. Of course this does not mean that K is of the form K(x,t) = .007 x t . Rather, we ask that K should be small in a sense developed below. The technique for getting a solution in this case is to iterate.

    Take phi 0(x) to be f(x) and phi 1 to be defined by

    It is reasonable to ask: does this generated sequence converge to a limit and in what sense does it converge? The answer to both questions can be found under appropriate hypothesis on K.

    Theorem XIV.3. If K satisfies the condition that

    max_x int(|K(x,t|, t=0..1) < 1                           (14.4)

    then limp phi p(x) exists and the convergence is uniform on [0,1] - in the sense that if u = limpphi p then

           limp maxx | u(x) - phi p(x) | = 0.

    Proof. Note that

    Furthermore, if p is a positive integer, the distance between successive iterates can be computed:

    Inductively, this does not exceed

    Thus, if

    and n > m then

    Hence, the sequence {phi p} of functions converges uniformly on [0,1] to a limit function and this limit provides a solution to the equation

    Corollary XIV.4. If


           u = limp phi p


    Sometimes it is convenient to express the iteration as an infinite series, called the Neumann series, i.e., the sum of psin = phin-phin-1. We reason this way in the next example.

    Model Problem XIV.5. Consider the integral equation

    u(x) = int(exp(t-x) u(t), t=0..x) + g(x),

    where g(x) is given. We wish to solve for u(x), and we try the method of iteration.

    We begin with the guess psi0 = g(x), and calculate the next couple of iterates:

    psi_1(x) = int(exp(t-x) g(t), t=0..x).
    psi_1(x) = int(exp(t-x) g(t), t=0..x).

    This integral can be simplified by reversing the order of integration. Setting the limits takes a moment of reflection, and may be helped by the following diagram:

    the triangle in the t-t<sub>1</sub>
plane such that 0 < t < t<sub>1</sub> < x

    The relationship of the variables is 0 < t < t1 < x. If the first (inside) integral is in the variable t, then it runs from 0 to t1, and then the second integral in the variable t1 runs from 0 to x. If we reverse the order, the first integral, in the variable t1, runs from t to x, and the second integral runs from 0 to x. We find that psi2(x) is:

    int(g(t) (x-t) exp(t-x) dt

    If we now calculate the further iterates, we find inductively that

    psi_n = int(g(t) ((x-t)^{n-1}/(n-1)!) 
exp(t-x) dt. The special feature about this example is that the iterates can be summed up, and when we recall that sum of (x-t)^n /n! = exp(x-t), we get: u(x) = g(x) + int(g(t), t=0..x. We leave it as an exercise to check this solution.

    It is a miracle when the series for K sums in closed form like this, but that is not important in applications, since the convergence of the Neumann series implies that we can calculate the answer to any desired accuracy.

    There is different, independent, way in which K can be considered small, which leads to convergence of the iteration process in the norm of L2[0,1]. This hypothesis asks that

    double integral of |K|^2 is finite                           (14.5)

    Theorem XIV.6. If K satisfies the Hilbert- Schmidt condition (14.4), then limp phip(x) exists and the convergence is in the r.m.s. sense, that is:

           limp || u(x) - phip(x) || = 0.

    INDICATION OF PROOF. The analysis of the nature of the convergence will go like this:

    ||phi1 - phi2|| 2

    is defined to be

    As a consequence, the sequence phin is Cauchy convergent:

    Let's state the conclusion in a careful way:

    Corollary XIV.7. If r := sqrt(int(|K|^2 dx dt),


    u = limp phip converges and

    ||u -phi<sub>m</sub>|| <u><</u> ||f|| r<sup>m+1</sup>/(1-r)
    Definition The resolvent of an operator K is the inverse operator (lambda - K)-1. This is same as the solution operator for the equation y = lambdaKy + f Often the number lambda is taken as 1, and unless stated otherwise we shall always do this.

    Before addressing the final case - where

  • K does not have a separable kernel,
  • The smallness condition (14.4) fails, and
  • The Hilbert-Schmidt smallness condition (14.5) fails,
  • we generate resolvents for the integral equations.

    Re-examining the iteration process:

        phi 0(x) = f(x),

        phi 1(x) = Kphi 0(x) + f(x)

        phi 2(x) = K(K(phi 0))x + K(f)(x) + f(x)

    One writes phi 0=f, phi 1=Kf+f, phi 2 = K[Kf+f] + f = K2f+Kf+f, .....

    In fact, with

    Hence, the kernel K2 associated with K2 is



    We have, in this section, conditions which imply that

    Sigmap=1 Kpf converges and that its limit y satisfies y = Ky + f. We have thus defined the resolvent for K,

    (1 - K)-1 = 1 + R, where
    R = Sigmap=1 Kp.                           (14.6)
    Note that R operates on elements of L2[0,1], and, subject to convergence, it has an integral kernel. The solution of y = Ky + f

    has a very similar form to the equation itself.  y(x) = [( 1 + <b>R</b> ) f](x) = f(x) + int(R(x,t) f(t), t=0..1)

    Indeed, f = - Rf + y. Algebraically, we can identifying (1 + R) with (1 - K ) -1 = 1 + K(1 - K)-1, so that R = K (1 - K)-1. Please refer to the accompanying Mathematica notebook or Maple worksheet for the solution by iteration of a typical integral equation, including error estimates.

    Continue with this chapter

    Link to

  • chapter XIII
  • Table of Contents
  • Evans Harrell's home page
  • Jim Herod's home page