next up previous contents index
Next: Singular Boundary Value Problem: Up: Boundary Value Problem via Previous: Eigenfunctions via Integral Equations   Contents   Index


Types of Integral Equations

It is evident that different types of boundary value problems give rise to different types of integral equations.

A.  Fredholm Equations

The inhomogeneous boundary value problem gave rise to Eq.(4.53), whose form is

$\displaystyle u(x)$ $\displaystyle =$ $\displaystyle \lambda\int^b_a K(x;\xi )u(\xi )d\xi +\varphi (x)\,.$ (455)

In this case, $ K(x;\xi )=G(x;\rho )\rho (\xi )$ and $ \varphi$ are known functions, and $ u$ is the unknown function.

The integration limits $ a$ and $ b$ are fixed. An integral equation for $ u(x)$ of the form Eq. (4.55) is called inhomogeneous Fredholm equation of the second kind. The expression $ K(x;\xi )$ is called the ``kernel'' of the integral equation.

A homogeneous Fredholm equation of the second kind is obtained by dropping the function $ \varphi (x)$ ,

$\displaystyle u(x)=\lambda\int^b_a K(x;\xi )u(\xi )d\xi\,.
$

Equation (4.54) and the subsequent eigenvalue equations are examples of such equations.

A Fredholm equation of the first kind has the form

$\displaystyle \varphi (x) = \int^b_a K(x;\xi )u(\xi )d\xi
$

whenever $ \varphi (x)$ is a known function and $ u(\xi)$ is the unknown function.

B.  Volterra Equations

Fredholm equations are based on definite integrals. If the integration limits are variable, then the corresponding integral equations are Volterra equations. An inhomogeneous Volterra equation of the second kind, corresponding to Eq. (4.55), has the form

$\displaystyle u(x)$ $\displaystyle =$ $\displaystyle \int^x_a K(x;\xi )u(\xi )d\xi +\varphi (\xi )\,.$ (456)

If $ \varphi =0$ , then one has a homogeneous Volterra equation of the second kind. By contrast, a Volterra equation of the first kind has the form

$\displaystyle \varphi (x)=\int^x_a K(x,\xi )u(\xi )d\xi\,,
$

where $ \varphi$ is known and $ u$ is the unknown function. A Volterra integral equation may be viewed as a Fredholm equation whose kernel vanishes for $ x<
\xi$ . Thus letting
$\displaystyle M (x;\xi )$ $\displaystyle =$ $\displaystyle 0 ~~\qquad \qquad~~~~x<\xi$  
$\displaystyle M (x;\xi )$ $\displaystyle =$ $\displaystyle K(x;\xi ) ~~\qquad~~\xi <x\,,$  

one finds that the Volterra Eq. (4.56) becomes

$\displaystyle u(x)=\int^b_a M(x;\xi )u(\xi )d\xi +\varphi (\xi )\,,
$

whose form is that of a Fredholm equation.

One of the prominent examples giving rise to Volterra's integral equations are initial value problems. To illustrate this point, consider the motion of a simple harmonic oscillator governed by the equation

$\displaystyle \frac{d^2u}{d\tau^2} +\omega^2 u$ $\displaystyle =$ 0 (457)

and the initial conditions
$\displaystyle u(0)$ $\displaystyle =$ $\displaystyle d$  
$\displaystyle \dot u (0)$ $\displaystyle =$ $\displaystyle e\,.$  

The Green's function for this problem is depicted in Figure 4.5 on page [*]. It is the response to the impulse $ \delta (t-\tau )$ , and it satisfies

$\displaystyle \frac{d^2G^R(t;\tau )}{dt^2}$ $\displaystyle =-\delta (t-\tau )\,;~~G^R(0;\tau )=0\,,~~ \left.\frac{dG^R(t;\tau )}{dt}\right\vert _{t=0} = 0$    

or


$\displaystyle \frac{d^2G^R(t;\tau )}{d\tau^2}$ $\displaystyle = -\delta (t-\tau )\,;~~G^R(t,\tau )=0\,, ~~\textrm{for~all}~~t<\tau\,,$ (458)

in spite of the fact that $ G^R(t;\tau )\not= G^R(\tau ;t)$ (Why? Hint: what is the second derivative of a function that depends only on $ t-\tau$ ?). To obtain the integral equation multiply Eq. (4.57) by $ G^R(t;\tau )$ and Eq. (4.58) by $ u(\tau )$ . One finds
$\displaystyle G^R(t;\tau )\left[ \frac{d^2u}{d\tau^2} +\omega^2\right] u$ $\displaystyle =$ 0  
$\displaystyle u(\tau)\frac{d^2G^R(t;\tau )}{d\tau^2}$ $\displaystyle =$ $\displaystyle -\delta (t-\tau )u(\tau)$  

Subtraction yields a l.h.s. whose second derivative terms consolidate into a total derivative (Lagrange's identity!):

$\displaystyle \frac{d}{d\tau}\left\{
G^R(t;\tau )\frac{du}{d\tau}-
u(\tau)\frac...
...au )}{d\tau} \right\}
+\omega^2 G^R(t;\tau )u(\tau)
= -\delta (t-\tau )u(\tau)
$

Next perform the integration $ \int^{t^+}_0 \cdots d\tau$ , where $ t^+$ signifies taking the limit of the integral as $ \tau \to
t$ from the side for which $ \tau >t$ . One obtains

$\displaystyle \int^{t^+}_0 d\tau\left\{ G^R(t;\tau )\frac{d^2u(\tau)}{d\tau^2} ...
...\tau )}{d\tau^2}\right\} +\omega^2\int^{t^+}_0 G^R(t,\tau )
u(\tau)d\tau =u(t)
$

or with the help of the property $ G^R(t;\tau )=0$ whenever $ t<\tau$ ,

$\displaystyle u(t) =\omega^2\int^{t^+}_0 G^R(t;\tau )u(\tau )d\tau +u(0)\left.\frac{dG^R
(t;\tau )}{d\tau}\right\vert _{\tau =0} -\dot u (0) G^R(t;0)\,.
$

Here $ u(0)$ and $ \dot u(0)$ are the initial amplitude and velocity of the simple harmonic oscillator, and they are now intrinsically incorporated in an inhomogeneous Volterra equation of the second kind. In this integral equation $ u(\tau )$ is the unknown function to be determined. However, the utility of this integral equation, which is based on the Green's function $ G^R(t;\tau )$ , is eclipsed by an integral equation which is based a another Green's function, say $ g^R(t;\tau)$ , satisfying

$\displaystyle \frac{d^2g^R(t;\tau )}{d\tau^2}+\omega^2 g^R(t;\tau )
= -\delta (t-\tau )\,;~~g^R(t,\tau )=0\,,
~~\textrm{for~all}~~t<\tau
$

similar to Eq.(4.57). Following the same derivation steps, one finds that the $ \omega^2$ -term gets cancelled.

$\displaystyle u(t) =u(0)\left.\frac{dg^R
(t;\tau )}{d\tau}\right\vert _{\tau =0} -\dot u (0) g^R(t;0)~.
$

The integral has diappeared. One is left with the solution to the problem one is actually trying to solve. The overall conclusion is this: Picking the right Green's function for the problem speeds up the process of reaching one's goal.

Exercise 49.1 (TRANSLATION INVARIANT INTEGRATION KERNEL)

Consider the inhomogeneous Fredholm equation of the second kind,

$\displaystyle u(x)=\lambda {1\over \sqrt{2\pi}}\int^\infty_{-\infty}
K(x;\xi)~u(\xi)~d\xi ~~+~~\phi(x).
$

Here $ \lambda $ is a parameter and $ \phi $ is a known and given function. So is the integration kernel $ K$ , which in this problem is given to be translation invariant, i.e. you should assume that $ K(x;\xi)=v(x-\xi)$ , where $ v$ is a given function whose Fourier transform

$\displaystyle \hat v(k)={1\over \sqrt{2\pi}}\int^\infty_{-\infty} e^{-ikx} v(x)~dx
$

exists. SOLVE the integral equation by finding the function $ u(x)$ in terms of what is given.

Exercise 49.2

Look up an integral equation of the 2nd kind, either of the Volterra or of the Fredholm type. Submit it and its solution.


Lecture 35



next up previous contents index
Next: Singular Boundary Value Problem: Up: Boundary Value Problem via Previous: Eigenfunctions via Integral Equations   Contents   Index
Ulrich Gerlach 2010-12-09