Masujima M. Applied Mathematical Methods in Theoretical Physics

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136 4 Integral Equations of the Fredholm Type

However, in order for the integral equation to make sense, we must require the

integral



∞

dyφ(y)

to converge. This requires

Re α<0,

which in turn requires

λ>0. (4.5.9)

Thus, for λ ≤ 0, we have no solution, and for λ>0, we have

φ(x) = C



α

) − (e

α

)



, (4.5.10)

with α

and α

given by Eq. (4.5.7).

Note that, in this case, we have a continuous spectrum of eigenvalues (λ>0) for

which the homogeneous problem has a solution. The reason why the eigenvalue is

not discrete is that K(x, y) is not square-integrable.

4.6

Fredholm Integral Equation with a Translation Kernel

Suppose x ∈ (−∞, +∞) and the kernel is translation invariant, i.e.,

K(x, y) = K(x − y). (4.6.1)

Then the inhomogeneous Fredholm integral equation of the second kind is given

φ(x) = f (x) +λ



+∞

−∞

K(x − y)φ(y)dy. (4.6.2)

Take the Fourier transform of both sides of Eq. (4.6.2) to ﬁnd

φ(k) =

f (k) +λ

K(k)

φ(k).

Solve for

φ(k)toﬁnd

φ(k) =

f (k)

1 −λ

K(k)

. (4.6.3)

4.6 Fredholm Integral Equation with a Translation Kernel 137

Solution φ(x) is provided by inverting the Fourier transform

φ(k) obtained above.

It seems so simple, but there are some subtleties involved in the inversion of

φ(k).

We present some general discussion of the inversion of the Fourier transform.

Suppose that the function F(x) has the asymptotic forms

F(x) ∼



as x →+∞ (a > 0),

as x →−∞ (b > a > 0).

(4.6.4)

Namely, F(x) grows exponentially as x →+∞and decays exponentially as x →

−∞. Then the Fourier transform

F(k) =



+∞

−∞

−ikx

F(x)dx (4.6.5)

exists as long as

−b < Im k < −a. (4.6.6)

This is because the integrand has the magnitude



−ikx

F(x)



∼



+a)x

as x →+∞,

+b)x

as x →−∞,

whereweset

k = k

+ ik

,withk

and k

real.

With

−b < k

< −a,

the magnitude of the integrand vanishes exponentially at both ends.

The inverse Fourier transformation then becomes

F(x) =

2π



+∞−iγ

−∞−iγ

ikx

F(k)dk with a <γ <b. (4.6.7)

Now if b = a such that F(x) ∼ e

for

→∞(a > 0), then the inversion contour

is on γ = a and the Fourier transform exists only for k

=−a.

Similarly if the function F(x) decays exponentially as x →+∞and grows as

x →−∞, we are able to continue deﬁning the Fourier transform and its inverse

by going in the upper half plane.

As an example, a function like

F(x) = e

−α

(4.6.8)

138 4 Integral Equations of the Fredholm Type

which decays exponentially as x →±∞, has a Fourier transform which exists and

is analytic in

−α<k

<α. (4.6.9)

Withthesequaliﬁcations,weshouldbeabletoinvert

φ(k)toobtainthesolution

to the inhomogeneous problem (4.6.2).

Now comes the homogeneous problem,

(x) = λ



+∞

−∞

K(x − y)φ

(y)dy. (4.6.10)

By Fourier transforming Eq. (4.6.10), we obtain

(1 −λ

K(k))

(k) = 0. (4.6.11)

If 1 −λ

K(k) has no zeros for all k,thenwehave

(k) = 0 ⇒ φ

(x) = 0, (4.6.12)

i.e., no nontrivial solution exists for the homogeneous problem. If, on the other

hand, 1 −λ

K(k) has a zero of order n at k = α,

(k) can be allowed to be of the

form

(k) = C

δ(k − α) + C

δ(k − α) +···+C





n−1

δ(k − α).

On inversion, we ﬁnd

(x) = C

iαx

+ C

iαx

+···+C

n−1

iαx

= e

iαx



j=1

j−1

. (4.6.13)

For the homogeneous problem, we choose the inversion contour of the Fourier

transform based on the asymptotic behavior of the kernel, a point to be discussed

in the following example.

 Example 4.3. Consider the homogeneous integral equation,

(x) = λ



+∞

−∞

−

x−y

(y)dy. (4.6.14)

Solution. Since the kernel vanishes exponentially as e

−y

as y →∞and as e

y →−∞, we need not require φ

(y)tovanishasy →±∞; rather, more generally

we may permit

(y) →



(1−ε)y

as y →∞,

(−1+ε)y

as y →−∞,

4.6 Fredholm Integral Equation with a Translation Kernel 139

and the integral equation still makes sense. So in the Fourier transform, we may

allow

−1 < k

< 1, (4.6.15)

and we still have a valid solution.

The Fourier transform of e

−α

is given by



+∞

−∞

−ikx

−α

dx = 2α



+ α



. (4.6.16)

Taking the Fourier transform of the homogeneous equation with α = 1, we ﬁnd

(k) = (2λ(k

+ 1))

(k),

from which, we obtain

+ 1 −2λ

(k) = 0.

So there exists no nontrivial solution unless k =±i

√

1 − 2λ. By the inversion

formula, φ

(x) is a superposition of e

+ikx

terms with amplitude

(k). But

(k)is

zero for all but k =±i

√

1 − 2λ. Hence we may conclude tentatively

(x) = C

−

√

1−2λx

+ C

√

1−2λx

However, we can at most allow φ

(x)togrowasfastase

as x →∞and as e

−x

x →−∞, as we discussed above. Thus further analysis is in order.

Case (1). 1 −2λ<0, or λ>12.

(x) is oscillatory and is given by

(x) = C

−i

√

2λ−1x

+ C

√

2λ−1x

. (4.6.17a)

Case (2). 0 < 1 −2λ<1, or 0 <λ<12.

(x) grows less fast than e

→∞:

(x) = C

−

√

1−2λx

+ C

√

1−2λx

. (4.6.17b)

Case (3). 1 −2λ ≥ 1, or λ ≤ 0.

No acceptable solution for φ

(x) exists, since e

√

1−2λx

grows faster than e

→∞.

Case (4). λ = 12.

(x) = C

x . (4.6.17c)

Now consider the corresponding inhomogeneous problem.

140 4 Integral Equations of the Fredholm Type

 Example 4.4. onsider the inhomogeneous integral equation,

φ(x) = ae

−α

+ λ



+∞

−∞

−

x−y

φ(y)dy. (4.6.18)

Solution. On taking the Fourier transform of Eq. (4.6.18), we obtain

φ(k) = (2aα (k

+ α

)) +(2λ(k

+ 1))

φ(k).

Solving for

φ(k), we obtain





2aα(k

+ 1)

+ 1 −2λ)(k

+ α

)

To invert the latter transform, we note that depending on whether λ is larger or

smaller than 12, the poles k =±

√

2λ − 1 lie on the real or imaginary axis of the

complex k plane. What we can do is to choose any contour for the inversion within

the strip

−min(α,1)< k

< min(α, 1) (4.6.19)

to get a particular solution to our equation and we may then add any multiple of

the homogeneous solution when the latter exists. The reason for choosing the strip

(4.6.19) instead of

−1 < k

< 1

in this case is that, in order for the Fourier transform of the inhomogeneous term

−α

to exist, we must also restrict our attention to

−α<k

<α.

Consider the ﬁrst three cases given in Example 4.3.

Cases (2) and (3). λ<12.

In these cases, we have 1 − 2λ>0. Hence





has simple poles at k =±iα and

k =±i

√

1 − 2λ. To ﬁnd a particular solution, use the real k-axis as the integration

contour for the inverse Fourier transformation. Then φ

(

)

is given by

(

)

2aα

2π



+∞

−∞

dke

ikx

+ 1)

+ 1 −2λ)(k

+ α

)

For x > 0, we close the contour in the upper half plane to obtain

(

)

1 −2λ − α





1 −α



−αx

−

2λα

√

1 −2λ

−

√

1−2λx



4.6 Fredholm Integral Equation with a Translation Kernel 141

For x < 0, we close the contour in the lower half plane to get an identical result

with x replaced with −x. Our particular solution φ

(

)

is given by

(

)

1 −2λ − α





1 −α



−α

−

2λα

√

1 −2λ

−

√

1−2λ



. (4.6.20)

For Case (3), this is the unique solution because there exists no acceptable

homogeneous solution, while for Case (2) we must also add the homogeneous part

given by

(

)

= C

−

√

1−2λx

+ C

√

1−2λx

Case (1). λ>12.

In this case, we have 1 −2λ<0. Hence





has simple poles at k =±iα and

k =±

√

2λ − 1. To do the inversion for the particular solution, we can take any

of the contours (1), (2), (3), or (4) as displayed in Figures 4.1–4.4, or Principal

Value contours which are equivalent to half the sum of the ﬁrst two contours

(1)

2l − 1

−

2l − 1

Fig. 4.1 The inversion contour (1) for Case (1).

−

2l − 1

−

(2)

Fig. 4.2 The inversion contour (2) for Case (1).

142 4 Integral Equations of the Fredholm Type

−

2l − 1

−

2l − 1

(3)

Fig. 4.3 The inversion contour (3) for Case (1).

−

2l − 1

−

(4)

Fig. 4.4 The inversion contour (4) for Case (1).

or half the sum of the latter two contours. Any of these differs by a multiple of

the homogeneous solution. Consider a particular choice (4) for the inversion. For

x > 0, we close the contour in the upper half plane. Then our particular solution

(

)

is given by

(

)

1 −2λ − α



(1 −α

−αx

2λαi

√

2λ − 1

−i

√

2λ−1x



For x < 0, we close the contour in the lower half plane to get an identical result

with x replaced with −x. So, in general, we can write our particular solution with

the inversion contour (4) as,

(

)

1 −2λ − α





1 −α



−α

2λαi

√

2λ − 1

−i

√

2λ−1



, (4.6.21)

4.8 Problems for Chapter 4 143

to which must be added the homogeneous part for Case (1) which reads

(

)

= C

−i

√

2λ−1x

+ C

√

2λ−1x

4.7

System of Fredholm Integral Equations of the Second Kind

We solve the system of Fredholm integral equations of the second kind,

(x) − λ





j=1

(x, y)φ

(y)dy = f

(x), i = 1, 2, ..., n , (4.7.1)

where the kernels K

(x, y) are square-integrable. We ﬁrst extend the basic interval

from [a, b]to[a, a + n(b − a)], and set

x + (i − 1)(b − a) = X < a + i(b − a), y + (j − 1)(b − a) = Y < a + j(b − a),

(4.7.2)

φ(X) = φ

(x), K(X, Y) = K

(x, y), f (X) = f

(x). (4.7.3)

We then obtain the Fredholm integral equation of the second kind,

φ(X) − λ



a+n(b−a)

K(X, Y)φ(Y)dY = f (X), (4.7.4)

where the kernel K(X, Y) is discontinuous in general but is square-integrable

on account of the square-integrability of K

(x, y). The solution φ(X) to Eq. (4.7.4)

provides the solutions φ

(x) to Eq. (4.7.1) with Eqs. (4.7.2) and (4.7.3).

4.8

Problems for Chapter 4

4.1. Calculate D(λ)for

(a) K(x, y) =



xy, y ≤ x,

0, otherwise.

(b) K(x, y) = xy,0≤ x, y ≤ 1.



g(x)h(y), y ≤ x,

0, otherwise.

(d) K(x, y) = g(x)h(y), 0 ≤ x, y ≤ 1.

Find zero of D(λ) for each case.

144 4 Integral Equations of the Fredholm Type

4.2. (due to H. C.). Solve

φ(x) = λ





φ(y)

(y + 1)







+∞

φ(y)

(y + 1)



, a > 0.

Find all eigenvalues and eigenfunctions.

4.3. (due to H. C.). Solve the Fredholm integral equation of the second kind,

given that

K(x − y) = e

−

x−y

, f (x) =



x , x > 0,

0, x < 0.

4.4. (due to H. C.). Solve the Fredholm integral equation of the second kind,

given that

K(x − y) = e

−

x−y

, f (x) = x for −∞< x < +∞.

4.5. (due to H. C.). Solve

φ(x) +λ



−1

K(x, y)φ(y)dy = 1, −1 ≤ x ≤ 1, K(x, y) =



1 −y

1 −x

Find all eigenvalues of K(x, y). Calculate also D(λ)andD(x, y;λ).

4.6. (due to H. C.). Solve the Fredholm integral equation of the second kind,

φ(x) = e

−

+λ



+∞

−∞

cosh(x − y)

φ(y)dy.

Hint:



+∞

−∞

ikx

cosh x

dx =

cosh(πk2)

4.7. (due to H. C. and D. M.). Consider the integral equation,

φ(x) = λ



+∞

−∞

√

2π

ixy

φ(y), −∞ < x < ∞.

(a) Show that there are only four eigenvalues of the kernel

(1

√

2π)exp[ixy]. What are these?

4.8 Problems for Chapter 4 145

(b) Show by an explicit calculation that the functions

(x) = exp[−x

2]H

(x),

where

(x) ≡ (−1)

exp[x

]

exp[−x

are Hermite polynomials, are eigenfunctions with the corresponding

eigenvalues (i)

,(n = 0, 1, 2, ...). Why should one expect φ

(x)tobe

Fourier transforms of themselves?

Hint: Think of the Schr

odinger equation for the harmonic oscillator.

(x)}

form a complete set, in

some sense, show that any square-integrable solution is of the form

φ(x) = f (x) +C

f (x),

where f (x) is an arbitrary even or odd, square-integrable function with the

Fourier transform

f (k), and C is a suitable constant. Evaluate C and relate

its values to the eigenvalues found in (a).

(d) From (c), construct a solution by taking f (x) = exp[−ax

2], a > 0.

4.8. (due to H. C.). Find an eigenvalue and the corresponding eigenfunction for

K(x, y) = exp[−(ax

+ 2bxy + cy

)], −∞ < x, y < ∞, a + c > 0.

4.9. Consider the homogeneous integral equation,

φ(x) = λ



∞

−∞

K(x, y)φ(y)dy, −∞ < x < ∞,

where

K(x, y) =

√

1 −t

exp[

+ y

]exp[−

− 2xyt

1 −t

], t ﬁxed,

0 < t < 1.

(a) Show directly that φ

(x) = exp[−x

2] is an eigenfunction of K(x, y)

corresponding to the eigenvalue λ = λ

= 1

√

π.