1 Introduction - California Institute of Technologyfcp/math/integralEquations/integral...2.2 Laplace...

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Physics 129b Integral Equations 051012 F. Porter Revision 150928 F. Porter 1 Introduction The integral equation problem is to find the solution to: h(x)f (x)= g(x)+ λ Z b a k(x, y)f (y)dy. (1) We are given functions h(x), g(x), k(x, y), and wish to determine f (x). The quantity λ is a parameter, which may be complex in general. The bivariate function k(x, y) is called the kernel of the integral equation. We shall assume that h(x) and g(x) are defined and continuous on the interval a x b, and that the kernel is defined and continuous on a x b and a y b. Here we will concentrate on the problem for real variables x and y. The functions may be complex-valued, although we will sometimes simplify the discussion by considering real functions. However, many of the results can be generalized in fairly obvious ways, such as relaxation to piece- wise continuous functions, and generalization to multiple dimensions. There are many resources for further reading on this subject. Some of the popular ones among physicists include the “classic” texts by Mathews and Walker, Courant and Hilbert, Whittaker and Watson, and Margenau and Murphy, as well as the newer texts by Arfken, and Riley, Hobson, and Bence. 2 Integral Transforms If h(x) = 0, we can take λ = -1 without loss of generality and obtain the integral equation: g(x)= Z b a k(x, y)f (y)dy. (2) This is called a Fredholm equation of the first kind or an integral transform. Particularly important examples of integral transforms include the Fourier transform and the Laplace transform, which we now discuss. 1

Transcript of 1 Introduction - California Institute of Technologyfcp/math/integralEquations/integral...2.2 Laplace...

Page 1: 1 Introduction - California Institute of Technologyfcp/math/integralEquations/integral...2.2 Laplace Transforms The Laplace transform is an integral transform of the form: F(s) = Z

Physics 129bIntegral Equations051012 F. Porter

Revision 150928 F. Porter

1 Introduction

The integral equation problem is to find the solution to:

h(x)f(x) = g(x) + λ∫ b

ak(x, y)f(y)dy. (1)

We are given functions h(x), g(x), k(x, y), and wish to determine f(x). Thequantity λ is a parameter, which may be complex in general. The bivariatefunction k(x, y) is called the kernel of the integral equation.

We shall assume that h(x) and g(x) are defined and continuous on theinterval a ≤ x ≤ b, and that the kernel is defined and continuous on a ≤ x ≤ band a ≤ y ≤ b. Here we will concentrate on the problem for real variablesx and y. The functions may be complex-valued, although we will sometimessimplify the discussion by considering real functions. However, many of theresults can be generalized in fairly obvious ways, such as relaxation to piece-wise continuous functions, and generalization to multiple dimensions.

There are many resources for further reading on this subject. Some ofthe popular ones among physicists include the “classic” texts by Mathewsand Walker, Courant and Hilbert, Whittaker and Watson, and Margenauand Murphy, as well as the newer texts by Arfken, and Riley, Hobson, andBence.

2 Integral Transforms

If h(x) = 0, we can take λ = −1 without loss of generality and obtain theintegral equation:

g(x) =∫ b

ak(x, y)f(y)dy. (2)

This is called a Fredholm equation of the first kind or an integraltransform. Particularly important examples of integral transforms includethe Fourier transform and the Laplace transform, which we now discuss.

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2.1 Fourier Transforms

A special case of a Fredholm equation of the first kind is

a = −∞ (3)

b = +∞ (4)

k(x, y) =1√2πe−ixy. (5)

This is known as the Fourier transform:

g(x) =1√2π

∫ ∞−∞

e−ixyf(y)dy (6)

Note that the kernel is complex in this case.The solution to this equation is given by:

f(y) =1√2π

∫ ∞−∞

eixyg(x)dx. (7)

We’ll forego rigor here and give the “physicist’s” demonstration of this:

g(x) =1

∫ ∞−∞

e−ixydy∫ ∞−∞

eix′yg(x′)dx′ (8)

=1

∫ ∞−∞

g(x′)dx′∫ ∞−∞

ei(x′−x)ydy (9)

=∫ ∞−∞

g(x′)δ(x− x′)dx′ (10)

= g(x). (11)

Here, we have used the fact that the Dirac “delta-function” may be written

δ(x) =1

∫ ∞−∞

eixydy. (12)

The reader is encouraged to demonstrate this, if s/he has not done so before.It is instructive to notice that the Fourier transform may be regarded as

a limit of the Fourier series. Let f(x) be expanded in a Fourier series in abox of size [−L/2, L/2]:

f(x) =∞∑

n=−∞ane

2πinx/L. (13)

We have chosen periodic boundary conditions here: f(L/2) = f(−L/2).The an expansion coefficients may be determined for any given f(x) using

the orthogonality relations:

1

L

∫ L/2

−L/2e2πinx/Le−2πimx/Ldx = δmn. (14)

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Hence,

an =1

L

∫ L/2

−L/2f(x)e−2πinx/Ldx. (15)

Now consider taking the limit as L → ∞. In this limit, the summationgoes over to a continuous integral. Let y = 2πn/L and g(y) = Lan/

√2π.

Then, using dn = (L/2π)dy,

f(x) = limL→∞

∞∑n=−∞

ane2πinx/L (16)

= limL→∞

∞∑n=−∞

√2π

Lg(y)eixy (17)

=1√2π

∫ ∞−∞

eixyg(y)dy. (18)

Furthermore:

g(y) =Lan√

2π=

1√2π

∫ ∞−∞

f(x)e−ixydx. (19)

We thus verify our earlier statements, including the δ-function equivalence,assuming our limit procedure is acceptable.

Suppose now that f(y) is an even function, f(−y) = f(y). Then,

g(x) =1√2π

[∫ 0

−∞e−ixyf(y)dy +

∫ ∞0

e−ixyf(y)dy]

(20)

=1√2π

∫ ∞0

[eixy + e−ixy

]f(y) dy (21)

=

√2

π

∫ ∞0

f(y) cosxy dy. (22)

This is known as the Fourier cosine transform. It may be observed thatthe transform g(x) will also be an even function, and the solution for f(y) is:

f(y) =

√2

π

∫ ∞0

g(x) cosxy dx. (23)

Similarly, if f(y) is an odd function, we have the Fourier sine trans-form:

g(x) =

√2

π

∫ ∞0

f(y) sinxy dy, (24)

where a factor of −i has been absorbed. The solution for f(y) is

f(y) =

√2

π

∫ ∞0

g(x) sinxy dx. (25)

Let us briefly make some observations concerning an approach to a morerigorous discussion. Later we shall see that if the kernel k(x, y) satisfies

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conditions such as square-integrability on [a, b] then convenient behavior isachieved for the solutions of the integral equation. However, in the presentcase, we not only have |a|, |b| → ∞, but the kernel eixy nowhere approacheszero. Thus, great care is required to ensure valid results.

We may deal with this difficult situation by starting with a set of functionswhich are themselves sufficiently well-behaved (e.g., approach zero rapidlyas |x| → ∞) that the behavior of the kernel is mitigated. For example, inquantum mechanics we may construct our Hilbert space of acceptable wavefunctions on R3 by starting with a set S of functions f(x) where:

1. f(x) ∈ C∞, that is f(x) is an infinitely differentiable complex-valuedfunction on R3.

2. lim|x|→∞ |x|nd(x) = 0, ∀n, where d(x) is any partial derivative of f .That is, f and its derivatives fall off faster than any power of |x|.

We could approach the proof of the Fourier inverse theorem with morerigor than our limit of a series as follows: First, consider that subset of Sconsisting of Gaussian functions. Argue that any function in S may be ap-proximated aribtrarily closely by a series of Gaussians. Then note that the Sfunctions form a pre-Hilbert space (also known as an Euclidean space). Addthe completion to get a Hilbert space, and show that the theorem remainsvalid.

The Fourier transform appears in many physical situations via its con-nection with waves, for example:

<eixy = cosxy. (26)

In electronics we use the Fourier transform to translate “time domain” prob-lems in terms of “frequency domain” problems, with xy → ωt. An LCRcircuit is just a complex impedance for a given frequency, hence the integral-differential time-domain problem is translated into an algebraic problem inthe frequency domain. In quantum mechanics the position-space wave func-tions are related to momenutm-space wave functions via the Fourier trans-form.

2.1.1 Example: RC circuit

Suppose we wish to determine the “output” voltage Vo(t) in the simple circuitof Fig. 1. The time domain problem requires solving the equation:

Vo(t) =1

R1C

∫ t

−∞Vi(t

′) dt′ − 1

C

(1

R1

+1

R2

) ∫ t

−∞Vo(t

′) dt′. (27)

This is an integral equation, which we will encounter in Section 5.2 as a“Volterra’s equation of the second kind”.

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R

RCV (t) V (t)i o

1

2

Figure 1: A simple RC circuit problem.

If Vi(t) is a sinusoid waveform of a fixed frequency (ω), the circuit elementsmay be replaced by complex impedances:

R1 → Z1 = R1 (28)

R2 → Z2 = R2 (29)

C → ZC =1

iωC. (30)

Then it is a simple matter to solve for Vo(t):

Vo(t) = Vi(t)1

1 + R1

R2(1 + iωR2C)

, (31)

if Vi(t) = sin(ωt+ φ), and where it is understood that the real part is to betaken.

Students usually learn how to obtain the result in Eqn. 31 long beforethey know about the Fourier transform. However, it is really the result inthe frequency domain according to the Fourier transform. That is:

Vo(ω) =1√2π

∫ ∞−∞

Vo(t)e−iωt dt (32)

= Vi(ω)1

1 + R1

R2(1 + iωR2C)

. (33)

We are here using the “hat” ( ) notation to indicate the integral transform ofthe unhatted function. The answer to the problem for general (not necessarilysinusoidal) input Vi(t) is then:

Vo(t) =1√2π

∫ ∞−∞

Vo(ω)eiωt dω (34)

=1√2π

∫ ∞−∞

Vi(ω)eiωt

1 + R1

R2(1 + iωR2C)

dω. (35)

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2.2 Laplace Transforms

The Laplace transform is an integral transform of the form:

F (s) =∫ ∞

0f(x)e−sxdx. (36)

The “solution” for f(x) is:

f(x) =1

2πi

∫ c+i∞

c−i∞F (s)esxds, (37)

where x > 0.This transform can be useful for some functions where the Fourier trans-

form does not exist. Problems at x → +∞ are removed by multiplying bye−cx, where c is a positive real number. Then the problem at −∞ is repairedby multiplying by the unit step function θ(x):

θ(x) ≡

1 if x > 0,1/2 if x = 0, and0 if x < 0.

(38)

Thus, we have

g(y) =∫ ∞−∞

f(x)θ(x)e−cxe−ixydx (39)

=∫ ∞

0f(x)e−cxe−ixydx, (40)

where we have by convention also absorbed the 1/√

2π.The inverse Fourier transform is just:

e−cxθ(x)f(x) =1

∫ ∞−∞

g(y)eixydy. (41)

If we let s = c+ iy and define F (s) ≡ g(y) at s = c+ iy, then

F (s) =∫ ∞

0f(x)e−sxdx, (42)

and

f(x)θ(x) =1

∫ ∞−∞

F (s)ex(c+iy)dy (43)

=1

2πi

∫ c+i∞

c−i∞F (s)exsds, (44)

which is the above-asserted result.We group together here some useful theorems for Fourier and Laplace

transforms: First define some notation. Let

(Ff) (y) = g(y) =1√2π

∫ ∞−∞

f(x)e−ixydx (45)

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be the Fourier transform of f , and

(Lf) (s) = F (s) =∫ ∞

0f(x)e−sxdx (46)

be the Laplace transform of f . Finally, let T (“transform”) stand for eitherF or L.

The reader should immediately verify the following properties:

1. Linearity: For functions f and g and complex numbers α, β,

T (αf + βg) = α(T f) + β(T g). (47)

2. Transform of derivatives: Integrate by parts to show that

(Ff ′)(y) = iy(Ff)(y), (48)

assuming f(x)→ 0 as x→ ±∞, and

(Lf ′)(y) = s(Lf)(s)− f(0), (49)

assuming f(x) → 0 as x → ∞ and defining f(0) ≡ limx→0+ f(x).The procedure here may be iterated to obtain expressions for higherderivatives.

3. Transform of integrals:{F[∫

f(x)dx]}

(y) =1

iy(Ff) (y) + Cδ(y), (50)

where C is an arbitrary constant arising from the arbitrary constant ofintegration of an indefinite integral;{

L[∫ x

0f(t)dt

]}(s) =

∫ ∞0

dxe−sx∫ x

0f(t)dt

=∫ ∞

0dtf(t)

∫ ∞t

dxe−sx

=1

s(Lf) (s). (51)

4. Translation:

[Ff(x+ a)] (y) = eiay [Ff ] (y), (52)

[Lf(x+ a)] (s) = eas [Lf ] (s)− θ(a)∫ a

0f(x)e−sxdx. (53)

5. Multiplication by an exponential:

{F [eaxf(x)]} (y) = (Ff) (y + ia), (54)

{L [eaxf(x)]} (s) = (Lf) (s− a). (55)

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6. Multiplication by x:

{F [xf(x)]} (y) = id

dy(Ff) (y), (56)

{L [xf(x)]} (s) = − d

ds(Lf) (s). (57)

An important notion that arises in applications of integral transforms isthat of a “convolution”:Definition (Convolution):Given two functions f1(x) and f2(x), and con-stants a, b, the convolution of f1 and f2 is:

g(x) =∫ b

af1(y)f2(x− y)dy. (58)

In the case of Fourier transforms, we are interested in −a = b = ∞. ForLaplace transforms, a = 0, b =∞. We then have the celebrated convolutiontheorem:

Theorem:

(Fg) (y) =√

2π (Ff1) (y) (Ff2) (y), (59)

(Lg) (y) =√

2π (Lf1) (y) (Lf2) (y). (60)

The proof is left as an exercise.

2.2.1 Laplace transform example: RC circuit

Let us return to the problem of determining the “output” voltage Vo(t) inthe simple circuit of Fig. 1. But now, suppose that we know that Vi(t) = 0for t < 0. In this case, the Laplace transform is an appropriate method totry.

There are, of course, many equivalent solution paths; let us think in termsof the currents: i(t) is the total current (through R1), iC(t) is the currentthrough the capacitor, and iR(t) is the current through R2. We know thati(t) = iC(t) + iR(t), and

Vo(t) = Vi(t)− i(t)R1 (61)

= [i(t)− iC(t)]R2 (62)

=Q

C=

1

C

∫ t

0iC(t′) dt′. (63)

This gives us three equations relating the unknowns Vo(t), i(t), and iC(t),which we could try to solve to obtain Vo(t). However, the integral in the lastequation complicates the solution. This is where the Laplace transform willhelp us.

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Corresponding to the first of the three equations we obtain (where thehat now indicates the Laplace transform):

Vo(s) =∫ ∞

0Vo(t)e

−st dt = Vi(s)− i(s)R1. (64)

Corresponding to the second, we have:

Vo(s) =[i(s)− iC(s)

]R2. (65)

For the third, we have:

Vo(s) =1

C

∫ ∞0

e−st∫ t

0iC(t′) dt′ dt

=1

C

∫ ∞0

iC(t′) dt′∫ ∞t′

e−st dt

=1

sCiC(s). (66)

Now we have three simultaneous algebraic equations, which may be readilysolved for Vo(s):

Vo(s) = Vi(s)1

1 + R1

R2(1 + sR2C)

. (67)

We note the similarity with Eqn. 33. Going back to the time domain, wefind:

Vo(t) =1

2πi

∫ a+i∞

a−i∞

1

1 + R1

R2(1 + sR2C)

Vi(s)est ds. (68)

For example, let’s suppose that Vi(t) is a brief pulse, V∆t ∼ A, at t = 0.Let’s model this as:

V (t) = Aδ(t− ε), (69)

where ε is a small positive number, inserted to make sure we don’t get intotrouble with the t = 0 boundary in the Laplace transform. Then:

Vi(s) =∫ ∞

0Aδ(t− ε)e−st dt = Ae−εs. (70)

Inserting into Eqn. 68, we have

Vo(t) =1

2πi

R2

R1 +R2

A∫ a+i∞

a−i∞

es(t−ε)

1 + τsds, (71)

where

τ ≡ R1R2

R1 +R2

C. (72)

The integrand has a pole at s = −1/τ . We thus choose the contour ofintegration as in Fig. 2. A contour of this form is known as a “Bromwich

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Re(s)

Im(s)

ax-1/ τ

Figure 2: The Bromwich contour, for the RC circuit problem.

contour”. In the limit R→∞ the integral around the semicircle is zero. Wethus have:

Vo(t) =1

2πi

R2

R1 +R2

A2πi(

Residue at − 1

τ

)

= AR2

R1 +R2

lims→−1/τ

(s+

1

τ

)es(t−ε)

1 + τs

= A1

R1Ce−tτ . (73)

In this simple problem, we could have guessed this result: At t = 0,we instantaneously put a voltage A/R1C on the capacitor. The time τ issimply the time constant for the capacitor to discharge through the parallel(R1, R2) combination. However, we may also treat more difficult problemswith this technique. The integral-differential equation in the t-domain be-comes a problem of finding the zeros of a polynomial in the s-domain, atwhich the residues are evaluated. This translated problem lends itself tonumerical solution.

3 Laplace’s Method for Ordinary Differential

Equations

There are many other integral transforms that we could investigate, but theFourier and Laplace transforms are the most ubiquitous in physics applica-tions. Rather than pursue other transforms, we’ll look at another example

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that suggests the breadth of application of these ideas. This is the “Laplace’sMethod” for the solution of ordinary differential equations. This method rep-resents a sort of generalization of the Laplace transform, using the feature ofturning derivatives into powers.

Suppose we wish to solve the differential equation:

n∑k=0

(ak + bkx)f (k)(x) = 0, (74)

where we use the notation

f (k)(x) ≡ dkf

dxk(x). (75)

In the spirit of the Laplace transform, assume a solution in the integralform

f(x) =∫CF (s)esx ds, (76)

where the contour will be chosen to suit the problem (not necessarily thecontour of the Laplace transform). We’ll insert this proposed solution intothe differential equation. Notice that

f ′(x) =∫CF (s)sesx ds. (77)

Thus,

0 =∫C

[U(s) + xV (s)]F (s)esx ds, (78)

where

U(s) =n∑k=0

aksk, (79)

V (s) =n∑k=0

bksk. (80)

We may eliminate the x in the xV term of Eqn. 78 with an integrationby parts:∫

CV (s)F (s)xesx ds = [V (s)F (s)esx]

∣∣∣∣∣c2

c1

−∫C

d

ds[V (s)F (s)] esx ds, (81)

where c1 and c2 are the endpoints of C. Hence

0 =∫C

{U(s)F (s)− d

ds[V (s)F (s)]

}esx ds+ [V (s)F (s)esx]

∣∣∣∣∣c2

c1

. (82)

We assume that we can choose C such that the integrated part vanishes.Then we will have a solution to the differential equation if

U(s)F (s)− d

ds[V (s)F (s)] = 0. (83)

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Note that we have transformed a problem with high-order derivatives (butonly first order polynomial coefficients) to a problem with first-order deriva-tives only, but with high-order polynomial coefficients.

Formally, we find a solution as:

d

ds[V (s)F (s)] = U(s)F (s) (84)

V (s)dF (s)

ds= U(s)F (s)− F (s)

dV (s)

ds(85)

d lnF

ds=

U

V− d lnV

ds(86)

lnF =∫ (

U

V− d lnV

ds

)ds (87)

=∫ U

Vds− lnV + lnA, (88)

where A is an arbitrary constant. Thus, the soluton for F (s) is;

F (s) =A

V (s)exp

[∫ s U(s′)

V (s′)ds′]. (89)

3.1 Example of Laplace’s method: Hermite Equation

The simple harmonic oscillator potential in the Schrodinger equation leadsto the Hermite differential equation:

f ′′(x)− 2xf ′(x) + 2νf(x) = 0, (90)

where ν is a constant. This is an equation that may lend itself to treatmentwith Laplace’s method; let’s try it.

First, we determine U(s) and V (s):

U(s) = a0 + a2s2 = 2ν + s2 (91)

V (s) = b1s = −2s. (92)

Substituting these into the general formula Eqn. 89, we have

F (s) ∝ − 1

2sexp

(−∫ s s2 + 2ν

2sds

)(93)

∝ − 1

2sexp

(−s

2

4− ν ln s

)(94)

∝ −e−s2/4

2sν+1. (95)

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C

Im(s)

Re(s)

Figure 3: A possible contour for the Hermite equation with non-integralconstant ν. The branch cut is along the negative real axis.

To find f(x) we substitute this result into Eqn. 76:

f(x) =∫CF (s)esx ds (96)

= A∫C

e−s2+2sx

sν+1ds, (97)

where A is an arbitrary constant, and where we have let s → 2s accordingto convention for this problem.

Now we are faced with the question of the choice of contour C. We atleast must require that the integrated part in Eqn. 82 vanish:

V (s)F (s)esx∣∣∣21∝ e−s

2+2sx

∣∣∣∣∣2

1

= 0. (98)

We’ll need to avoid s = 0 on our contour. If ν = n is a non-negative integer,we can take a circle around the origin, since the integrand is then analyticand single-valued everywhere except at s = 0. If ν 6= n, then s = 0 is abranch point, and we cannot choose C to circle the origin. We could in thiscase take C to be the contour of Fig. 3.

Let’s consider further the case with ν = n = 0, 1, 2, . . . Take C to be acircle around the origin. Pick by convention A = n!/2πi, and define:

Hn(x) ≡ n!

2πi

∫C

e−s2+2sx

sn+1ds. (99)

This is a powerful integral form for the “Hermite polynomials” (or, Hermitefunctions in general with the branch cut contour of Fig. 3). For example,

Hn(x) = 2πin!

2πi× residue at s = 0 of

(e−s

2+2sx

sn+1

). (100)

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Recall that the residue is the coefficient of the 1/s term in the Laurent seriesexpansion. Hence,

Hn(x) = n!×(coefficient of sn in e−s

2+2sx). (101)

That is,

e−s2+2sx =

∞∑n=0

Hn(x)

n!sn. (102)

This is the “generating function” for the Hermite polynomials.The term “generating function” is appropriate, since we have:

Hn(x) = lims→0

dn

dsne−s

2+2sx (103)

H0(x) = lims→0

e−s2+2sx = 1 (104)

H1(x) = lims→0

(−2s+ 2x)e−s2+2sx = 2x, (105)

and so forth.

4 Integral Equations of the Second Kind

Referring back to Eq. 1, if h(x) 6= 0 for a ≤ x ≤ b we may rewrite theproblem in a form with h(x) = 1:

g(x) = f(x)− λ∫ b

ak(x, y)f(y)dy. (106)

This is referred to as a linear integral equation of the second kind or as aFredholm equation of the second kind. It defines a linear transformationfrom function f to function g. To see this, let us denote this transformationby the letter L:

g(x) = (Lf)(x) = f(x)− λ∫ b

ak(x, y)f(y)dy. (107)

If

Lf1 = g1 and (108)

Lf2 = g2, (109)

then for aribtrary complex constants c1 and c2:

L(c1f1 + c2f2) = c1g1 + c2g2. (110)

Notice that we may sometimes find it convenient to use the notation:

|g〉 = L|f〉 = |f〉 − λK|f〉, (111)

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where K|f〉 indicates here the integral∫ ba k(x, y)f(y)dy. Our linear operator

is then written:L = I − λK, (112)

where I is the identity operator.We are interested in the problem of inverting this linear transformation

– given g, what is f? As it is a linear transformation, it should not besurprising that the techniques are analogous with those familiar in matrixequations. The difference is that we are now dealing with vector spaces thatare infinite-dimensional function spaces.

4.1 Homogeneous Equation, Eigenfunctions

It is especially useful in approaching this problem to first consider the specialcase g(x) = 0:

f(x) = λ∫ b

ak(x, y)f(y)dy. (113)

This is called the homogeneous integral equation. It has a trivial solutionf(x) = 0 for a ≤ x ≤ b.

If there exists a non-trivial solution f(x) to the homogeneous equation,then cf(x) is also a solution, and we may assume that our solution is “nor-malized” (at least up to some here-neglected questions of rigor related to thespecification of our function space):∫ b

a|f(x)|2dx = 1. (114)

If there are several solutions, f1, f2, f3, . . . , fn, then any linear combina-tion of these is also a solution. Hence, if we have several linearly independentsolutions, we can assume that they are orthogonal and normalized. If theyare not, we may use the Gram-Schmidt process to obtain such a set of or-thonormal solutions. We therefore assume, without loss of generality, that:∫ b

af ∗i (x)fj(x)dx = δij. (115)

Alternatively, we may use the familiar shorthand:

〈fi|fj〉 = δij, (116)

or even|f〉〈f | = If , (117)

where If is the identity matrix in the subspace spanned by {f}.A value of λ for which the homogeneous equation has non-trivial solutions

is called an eigenvalue of the equation (or, of the kernel). Note that the useof the term eigenvalue here is analogous with, but different in detail from the

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usage in matrices – our present eigenvalue is more similar with the inverse ofa matrix eigenvalue. The corresponding solutions are called eigenfunctionsof the kernel for eigenvalue λ. We have the following:

Theorem: There are a finite number of eigenfunctions fi corresponding toa given eigenvalue λ.

Proof: We’ll prove this for real functions, leaving the complex case as anexercise. Given an eigenfunction fj corresponding to eigenvalue λ, let:

pj(x) ≡∫ b

ak(x, y)fj(y)dy =

1

λfj(x). (118)

Now consider, for some set of n eigenfunctions corresponding to eigenvalueλ:

D(x) ≡ λ2∫ b

a

k(x, y)−n∑j=1

pj(x)fj(y)

2

dy. (119)

It must be that D(x) ≥ 0 because the integrand is nowhere negative for anyx. Note that the sum term may be regarded as an approximation to thekernel, hence D(x) is a measure of the closeness of the approximation. Withsome manipulation:

D(x) = λ2∫ b

a[k(x, y)]2 dy − 2λ2

∫ b

a

n∑j=1

k(x, y)pj(x)fj(y)dy

+λ2∫ b

a

n∑j=1

pj(x)fj(y)

2

dy

= λ2∫ b

a[k(x, y)]2 dy − 2λ2

n∑j=1

[pj(x)]2

+λ2n∑j=1

pj(x)n∑k=1

pk(x)∫ b

afj(y)fk(y)dy

= λ2∫ b

a[k(x, y)]2 dy − λ2

n∑j=1

[pj(x)]2 . (120)

With D(x) ≥ 0, we have thus proved a form of Bessel’s inequality. We mayrewrite the inequality as:

λ2∫ b

a[k(x, y)]2 dy ≥

n∑j=1

[fj(x)]2 . (121)

If we integrate both sides over x, we obtain:

λ2∫ b

a

∫ b

a[k(x, y)]2 dydx ≥

n∑j=1

∫ b

a[fj(x)]2 dx

≥ n, (122)

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using the normalization of the fj. As long as∫ ∫

k2dxdy is bounded, we seethat n must be finite. For finite a and b, this is certainly satisfied, by ourcontinuity assumption for k. Otherwise, we may impose this as a requirementon the kernel.

More generally, we regard “nice” kernels as those for which∫ b

a

∫ b

a[k(x, y)]2 dydx < ∞, (123)∫ b

a[k(x, y)]2 dx < U1, ∀y ∈ [a, b], (124)∫ b

a[k(x, y)]2 dy < U2, ∀x ∈ [a, b], (125)

where U1 and U2 are some fixed upper bounds. We will assume that theseconditions are satisfied in our following discussion. Note that the kernel mayactually be discontinuous and even become infinite in [a, b], as long as theseconditions are satisfied.

4.2 Degenerate Kernels

Definition (Degenerate Kernel):If we can write the kernel in the form:

k(x, y) =n∑i=1

φi(x)ψ∗i (y) (126)

(or K =∑ni=1 |φi〉〈ψi|), then the kernel is called degenerate. We may

assume that the φi(x) are linearly independent. Otherwise we could reducethe number of terms in the sum to use only independent functions. Likewisewe may assume that the ψi(x) are linearly independent.

The notion of a degenerate kernel is important due to two facts:

1. Any continuous function k(x, y) can be uniformly approximated bypolynomials in a closed interval. That is, the polynomials are “com-plete” on a closed bounded interval.

2. The solution of the integral equation for degenerate kernels is easy (atleast formally).

The first fact is known under the label Weierstrass ApproximationTheorem. A proof by construction may be found in Courant and Hilbert.We remind the reader of the notion of uniform convergence in the sense usedhere:Definition (Uniform Convergence):If S(z) =

∑∞n=0 un(z) and SN =

∑Nn=0 un(z),

then S(z) is said to be uniformly convergent over the set of points A ={z|z ∈ A} if, given any ε > 0, there exists an integer N such that

|S(z)− SN+k(z)| < ε, ∀k = 0, 1, 2, . . . and ∀z ∈ A. (127)

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Note that this is a rather strong form of convergence – a series may convergefor all z ∈ A, but may not be uniformly convergent.

Let us now pursue the second fact asserted above. We wish to solve forf :

g(x) = f(x)− λ∫ b

ak(x, y)f(y)dy. (128)

If the kernel is degenerate, we have:

g(x) = f(x)− λn∑i=1

φi(x)∫ b

aψ∗i (y)f(y)dy. (129)

We define the numbers:

gi ≡∫ b

aψ∗i (x)g(x)dx (130)

fi ≡∫ b

aψ∗i (x)f(x)dx (131)

cij ≡∫ b

aψ∗i (x)φj(x)dx. (132)

Multiply Eq. 128 through by ψ∗j (x) and integrate over x to obtain:

gj = fj − λn∑i=1

cjifi. (133)

This is a system of n linear equations in the n unknowns fi. Suppose thatthere is a unique solution f1, f2, . . . , fn to this system. It is readily verifiedthat a solution to the integral equation is:

f(x) = g(x) + λn∑i=1

fiφi(x). (134)

Substituting in:

g(x) = g(x) + λn∑i=1

fiφi(x)− λ∑i=1n

φi(x)∫ b

aψ∗i (y)

g(y) + λn∑j=1

fjφj(y)

dy= g(x) + λ

n∑i=1

φi(x)

fi −∫ b

aψ∗i (y)

g(y) + λn∑j=1

fjφj(y)

dy

= g(x) + λn∑i=1

φi(x)

fi −gi + λ

n∑j=1

cijfj

= g(x). (135)

Let us try an explicit example to illustrate how things work. We wish tosolve the equation:

x2 = f(x)− λ∫ 1

0x(1 + y)f(y)dy (136)

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In this case, n = 1, and it is clear that the solution is simply a quadraticpolynomial which can be determined directly. However, let us apply our newmethod instead. We have g(x) = x2 and k(x, y) = x(1 + y). The kernel isdegenerate, with φ1(x) = x and ψ1(y) = 1 + y. Our constants evaluate to:

g1 =∫ 1

0(1 + x)x2dx =

7

12(137)

c11 =∫ 1

0x(1 + x)dx =

5

6. (138)

The linear equation we need to solve is then:

7

12= f1 − λ

5

6f1, (139)

giving

f1 =7

2

1

6− 5λ, (140)

and

f(x) = x2 +7

2

λ

6− 5λx. (141)

The reader is encouraged to check that this is a solution to the originalequation, and that no solution exists if λ = 6/5.

To investigate this special value λ = 6/5, consider the homogeneous equa-tion:

f(x) = λ∫ 1

0x(1 + y)f(y)dy. (142)

We may use the same procedure in this case, except now g1 = 0 and we findthat

f1

(1− λ5

6

)= 0. (143)

Either f1 = 0 or λ = 6/5. If f1 = 0, then f(x) = g(x) + λf1φ1(x) = 0. Ifλ 6= 6/5 the only solution to the homogeneous equation is the trivial one. Butif λ = 6/5 the solution to the homogeneous equation is f(x) = ax, where ais arbitrary. The value λ = 6/5 is an (in this case the only) eigenvalue of theintegral equation, with corresponding normalized eigenfunction f(x) =

√3x.

This example suggests the plausibility of the important theorem in thenext section.

4.3 Fredholm Alternative Theorem

Theorem: Either the integral equation

f(x) = g(x) + λ∫ b

ak(x, y)f(y)dy, (144)

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with given λ, possesses a unique continuous solution f(x) for each con-tinuous function g(x) (and in particular f(x) = 0 if g(x) = 0), or theassociated homogeneous equation

f(x) = λ∫ b

ak(x, y)f(y)dy (145)

possesses a finite number of linearly independent solutions.

We’ll give an abbreviated proof of this theorem to establish the ideas; thereader may wish to fill in the rigorous details.

We have already demonstrated that there exists at most a finite numberof linearly independent solutions to the homogeneous equation. A good ap-proach to proving the remainder of the theorem is to first prove it for thecase of degenerate kernels. We’ll use the Dirac notation for this, suggestingthe applicability for linear operators in general. Thus, let

K =n∑i=1

|φi〉〈ψi| (146)

|f〉 = |g〉+ λn∑i=1

|φi〉〈ψi|f〉, (147)

and let

gi ≡ 〈ψi|g〉 (148)

fi ≡ 〈ψi|f〉 (149)

cij ≡ 〈ψj|φi〉. (150)

Then,

fj = gj + λn∑i=1

cjifi, (151)

or

g =

g1

g2...gn

= (I − λC) f , (152)

where C is the matrix formed of the cij constants.Thus, we have a system of n linear equations for the n unknowns {fi}.

Either the matrix I − λC is non-singular, in which case a unique solution fexists for any given g (in particular f = 0 if g = 0), or I − λC is singular, inwhich case the homogeneous equation f = λCf possesses a finite number oflinearly independent solutions. Up to some further considerations concerningcontinuity, this proves the theorem for the case of a degenerate kernel.

We may extend the proof to arbitrary kernels by appealing to the factthat any continuous funciton k(x, y) may be uniformly approximated by de-generate kernels in a closed interval (for example, see Courant and Hilbert).

There is an additional useful theorem under Fredholm’s name:

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Theorem: If the integral equation:

f(x) = g(x) + λ∫ b

ak(x, y)f(y)dy (153)

for given λ possesses a unique continuous solution for each continuousg(x), then the transposed equation:

t(x) = g(x) + λ∫ b

ak(y, x)t(y)dy (154)

also possesses a unique solution for each g. In the other case, ifthe homogeneous equation possesses n linearly independent solutions{f1, f2, . . . , fn}, then the transposed homogeneous equation

t(x) = λ∫ b

ak(y, x)t(y)dy (155)

also has n linearly independent solutions {t1, t2, . . . , tn}. In this case,the original inhomogeneous equation 153 has a solution if and only ifg(x) satisfies conditions:

〈g|ti〉 =∫ b

ag∗(x)ti(x)dx = 0, i = 1, 2, . . . , n. (156)

That is, g must be orthogonal to all of the eigenvectors of the transposedhomogeneous equation. Furthermore, in this case, the solution is onlydetermined up to addition of an arbitrary linear combination of theform:

c1f1 + c2f2 + . . .+ cnfn. (157)

Again, a promising approach to proving this is to first consider the case ofdegenerate kernels, and then generalize to arbitrary kernels.

5 Practical Approaches

We turn now to a discussion of some practical “tools of the trade” for solvingintegral equations.

5.1 Degenerate Kernels

If the kernel is degenerate, we have shown that the solution may be obtainedby transforming the problem to that of solving a system of linear equations.

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5.2 Volterra’s Equations

Integral equations of the form:

g(x) = λ∫ x

ak(x, y)f(y)dy (158)

f(x) = g(x) + λ∫ x

ak(x, y)f(y)dy (159)

are called Volterra’s equations of the first and second kind, respectively.One situation where such equations arise is when k(x, y) = 0 for y > x:k(x, y) = θ(x− y)`(x, y). Thus,∫ b

ak(x, y)f(y)dy =

∫ x

a`(x, y)f(y)dy. (160)

Consider Volterra’s equation of the first kind. Recall the fundamentaltheorem:

d

dx

∫ b(x)

a(x)f(y, x)dy = f(b, x)

db

dx− f(a, x)

da

dx+∫ b

a

∂f

∂x(y, x)dy. (161)

We may use this to transform the equation of the first kind to:

dg

dx(x) = λk(x, x)f(x) + λ

∫ x

a

∂k

∂x(x, y)f(y)dy. (162)

This is now a Volterra’s equation of the second kind, and the approach tosolution may thus be similar.

Notice that if the kernel is independent of x, k(x, y) = k(y), then thesolution to the equation of the first kind is simply:

f(x) =1

λk(x)

dg

dx(x). (163)

Let us try a simple example. Suppose we wish to find f(x) in:

x2 = 1 + λ∫ x

1xyf(y)dy. (164)

This may be solved with various approaches. Let φ(x) ≡∫ x

1 yf(y)dy. Then

φ(x) =x2 − 1

λx. (165)

Now take the derivative of both sides of the original equation:

2x = λx2f(x) + λ∫ x

1yf(y)dy = λx2f(x) + λφ(x). (166)

A bit of further algebra yields the answer:

f(x) =1

λx2

(x+

1

x

). (167)

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As always, especially when you have taken derivatives, it should be checkedthat the result actually solves the original equation!

This was pretty easy, but it is even easier if we notice that this problemis actually equivalent to one with an x-independent kernel. That is, we mayrewrite the equation as:

x2 − 1

x= λ

∫ x

1yf(y)dy. (168)

Then we may use Eq. 163 to obtain the solution.

5.2.1 Numerical Solution of Volterra’s equation

The Volterra’s equation readily lends itself to a numerical approach to solu-tion on a grid (or “mesh” or “lattice”). We note first that (absorbing thefactor λ into the definition of k for convenience):

f(a) = g(a) +∫ x=a

ak(a, y)f(y)dy

= g(a). (169)

This suggests building up a solution at arbitrary x by stepping along a gridstarting at x = a.

To carry out this program, we start by dividing the interval (a, x) into Nsteps, and define:

xn = a+ n∆, n = 0, 1, . . . , N, ∆ ≡ x− aN

. (170)

We have here defined a uniform grid, but that is not a requirement. Now let

gn = g(xn) (171)

fn = f(xn) (172)

knm = k(xn, xm). (173)

Note that f0 = g0.We may pick various approaches to the numerical integration, for exam-

ple, the trapezoidal rule gives:

∫ xn

ak(xn, y)f(y) dy ≈ ∆

(1

2kn0f0 +

n−1∑m=1

knmfm +1

2knnfn

). (174)

Substituting this into the Volterra equation yields, at x = xn:

fn = gn + ∆

(1

2kn0f0 +

n−1∑m=1

knmfm +1

2knnfn

), n = 1, 2, . . . N. (175)

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Solving for fn then gives:

fn =gn + ∆

(12kn0f0 +

∑n−1m=1 knmfm

)1− ∆

2knn

, n = 1, 2, . . . N. (176)

For example,

f0 = g0, (177)

f1 =g1 + ∆

2k10f0

1− ∆2k11

, (178)

f2 =g2 + ∆

2k20f0 + ∆k21f1

1− ∆2k22

, (179)

and so forth.We note that we don’t even have to explicitly solve a system of linear

equations, as we did for Fredholm’s equation with a degenerate kernel. Thereare of order

O

(N∑n=1

n

)= O

(N2)

(180)

operations in this algorithm. The accuracy may be estimated by looking atthe change as additional grid points are added.

5.3 Neumann Series Solution

Often an exact closed solution is elusive, and we resort to approximate meth-ods. For example, one common approach is the iterative solution. We startwith the integral equation:

f(x) = g(x) + λ∫ b

ak(x, y)f(y)dy. (181)

The iterative approach begins with setting

f1(x) = g(x). (182)

Substituting this into the integrand in the original equation gives:

f2(x) = g(x) + λ∫ b

ak(x, y)g(y)dy. (183)

Substituting this yields:

f3(x) = g(x) + λ∫ b

ak(x, y)

[g(y) + λ

∫ b

ak(y, y′)g(y′)dy′

]dy. (184)

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This may be continuted indefinitely, with the nth iterative solution given interms of the (n− 1)th:

fn(x) = g(x) + λ∫ b

ak(x, y)fn−1(y)dy (185)

= g(x) + λ∫ b

ak(x, y)g(y)dy (186)

+λ2∫ b

a

∫ b

ak(x, y)k(y, y′)g(y′)dydy′

+ . . .

+λn−1∫ b

a· · ·

∫ b

ak(x, y) · · · k(y(n−2)′, y(n−1)′)g(y(n−1)′)dy . . . dy(n−1)′.

If the method converges, then

f(x) = limn→∞

fn(x). (187)

This method is only useful if the series converges, and the faster the bet-ter. It will converge if the kernel is bounded and lambda is “small enough”.We won’t pursue this further here, except to note what happens if∫ b

ak(x, y)g(y)dy = 0. (188)

In this case, the series clearly converges, onto solution f(x) = g(x). However,this solution is not necessarily unique, as we may add any linear combinationof solutions to the homogeneous equation.

5.4 Fredholm Series

Better convergence properties are obtained with the Fredholm series. Asbefore, we wish to solve

f(x) = g(x) + λ∫ b

ak(x, y)f(y)dy. (189)

Let

D(λ) = 1− λ∫ b

ak(x, x)dx+

λ2

2!

∫ ∫ ∣∣∣∣ k(x, x) k(x, x′)k(x′, x) k(x′, x′)

∣∣∣∣ dxdx′−λ

3

3!

∫ ∫ ∫ ∣∣∣∣∣∣∣k(x, x) k(x, x′) k(x, x′′)k(x′, x) k(x′, x′) k(x′, x′′)k(x′′, x) k(x′′, x′) k(x′′, x′′)

∣∣∣∣∣∣∣ dxdx′dx′′+ . . . , (190)

and let

D(x, y;λ) = λk(x, y)− λ2∫ ∣∣∣∣ k(x, y) k(x, z)

k(z, y) k(z, z)

∣∣∣∣ dz

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+λ3

2!

∫ ∫ ∣∣∣∣∣∣∣k(x, y) k(x, z) k(x, z′)k(z, y) k(z, z) k(z, z′)k(z′, y) k(z′, z) k(z′, z′)

∣∣∣∣∣∣∣ dzdz′+ . . . , (191)

Note that not everyone uses the same convention for this notation. Forexample, Mathews and Walker defines D(x, y;λ) to be 1/λ times the quantitydefined here.

We have the following:

Theorem: If D(λ) 6= 0 and if the Fredholm’s equation has a solution, thenthe solution is, uniquely:

f(x) = g(x) +∫ b

a

D(x, y;λ)

D(λ)g(y)dy. (192)

The homogeneous equation f(x) = λ∫ ba k(x, y)f(y)dy has no continu-

ous non-trivial solutions unless D(λ) = 0.

A proof of this theorem may be found in Whittaker and Watson. Theproof may be approached as follows: Divide the range a < x < b into equalintervals and replace the original integral by a sum:

f(x) = g(x) + λn∑i=1

k(x, xi)f(xi)δ, (193)

where δ is the width of an interval, and xi is a value of x within intervali. This provides a system of linear equations for f(xi), which we may solveand take the limit as n → ∞, δ → 0. In this limit, D(λ) is the limit of thedeterminant matrix of coefficients expanded in powers of λ.

While the Fredholm series is cumbersome, it has the advantage over theNeumann series that the series D(λ) and D(x, y;λ) are guaranteed to con-verge.

There is a nice graphical representation of the Fredholm series; I’ll de-scribe a variant here. We let a line segment or smooth arc represent thekernel k(x, y), one end of the segment corresponds to variable x and theother end to variable y. If the segment closes on itself smoothly (e.g., wehave a circle), then the variables at the two “ends” are the same – we havek(x, x). The product of two kernels is represented by making two segmentsmeet at a point. The meeting ends correspond to the same variable, the firstvariable in one kernel and the second variable in the other kernel. One maythink of the lines as directed, such that the second variable, say, is at the“starting” end, and the first variable is at the “finishing” end. When two seg-ments meet, it is always that the “finish” of one is connected to the “start”of the other. We could draw arrows to keep track of this, but it actuallyisn’t needed in this application, since in practice we’ll always integrate over

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the repeated variables. A heavy dot on a segment breaks the segment intotwo meeting segments, according to the above rule, and furthermore meansintegration over the repeated variable with a factor of λ. For illustration ofthese rules:

k(x, y)

k(x, x)

k(x, y)k(y, z)

• λ∫k(x, y)k(y, z)dy

••• λ∫k(x, x)dx

Thus,

D(x, y;λ)

λ= −

(•••−•

)+

1

2!

(••• •••− • •• • + ••−• ••• + ••−• •••

)− . . . , (194)

and

D(λ) = 1− ••• +1

2!

(••• •••− • •• •

)

− 1

3!

(••• ••• •••− •••• •• • + • •

•• •• − •••• •• • + • •

•• •• − •••• •• •

)+ . . . (195)

Let us try a very simple example to see how things work. Suppose wewish to solve:

f(x) = x+ λ∫ 1

0xyf(y)dy. (196)

Of course, this may be readily solved by elementary means, but let us applyour new techniques. We have:

= k(x, y) = xy (197)

••• = λ∫ 1

0k(x, x)dx =

λ

3(198)

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• = λ∫ 1

0k(x, y)k(y, z)dy =

λ

3xz = ••• (199)

• •• • = λ2∫ 1

0

∫ 1

0k(x, y)k(y, x)dxdy =

3

)2

=(

•••)2. (200)

We thus notice that

••••••ndots = λn

∫ 1

0· · ·

∫ 1

0dx . . . dx(n)x2 . . .

[x(n)

]2

=(

•••)n. (201)

We may likewise show that

••• n

dots =(

•••)n. (202)

We find from this that all determinants of dimension ≥ 2 vanish. Wehave

D(λ) = 1− ••• = 1− λ

3(203)

1

λD(x, y;λ) = = xy. (204)

The solution in terms of the Fredholm series is then:

f(x) = g(x) +∫ 1

0

D(x, y;λ)

D(λ)g(y)dy

= x+3λ

3− λ

∫ 1

0xy2dy

=3

3− λx. (205)

Generalizing from this example, we remark that if the kernel is degenerate,

k(x, y) =n∑i=1

φi(x)ψi(y), (206)

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then D(λ) and D(x, y;λ) are polynomials of degree n in λ. The readeris invited to attempt a graphical “proof” of this. This provides anotheralgorithm for solving the degenerate kernel problem.

Now suppose that we attempt to solve our example with a Neumannseries. We have

f(x) = g(x) + λ∫k(x, y)g(y)dy + λ2

∫ ∫k(x, y)k(y, y′)g(y′)dydy′ + . . .

= x+ λx∫ 1

0y2dy + λ2x

∫ 1

0

∫ 1

0y2(y′)2dydy′ + . . .

= x∞∑n=0

3

)n. (207)

This series converges for |λ| < 3 to

f(x) =3

3− λx. (208)

This is the same result as the Fredholm solution above. However, the Neu-mann solution is only valid for |λ| < 3, while the Fredholm solution is validfor all λ 6= 3. At eigenvalue λ = 3, D(λ = 3) = 0.

At λ = 3, we expect a non-trivial solution to the homogeneous equation

f(x) = 3∫ 1

0xyf(y)dy. (209)

Indeed, f(x) = Ax solves this equation. The roots ofD(λ) are the eigenvaluesof the kernel. If the kernel is degenerate we only have a finite number ofeigenvalues.

6 Symmetric Kernels

Definition: If k(x, y) = k(y, x) then the kernel is called symmetric. Ifk(x, y) = k∗(y, x) then the kernel is called Hermitian.

Note that a real, Hermitian kernel is symmetric. For simplicity, we’ll restrictourselves to real symmetric kernels here,1 but the generalization to Hermitiankernels is readily accomplished (indeed is already done when we use Dirac’snotation). The study of such kernels via eigenfunctions is referred to as“Schmidt-Hilbert theory”. We will assume that our kernels are bounded inthe sense: ∫ b

a[k(x, y)]2dy ≤ M, (210)∫ b

a

[∂k

∂x(x, y)

]2

dy ≤ M ′, (211)

1Note that, since we are assuming real functions in this section, we do not put a complexconjugate in our scalar products. But don’t forget to put in the complex conjugate if youhave a problem with complex functions!

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where M and M ′ are finite.Our approach to studying the symmetric kernel problem will be to analyze

it in terms of the solutions to the homogeneous equation. We have thefollowing:

Theorem: Every continuous symmetric kernel (not identically zero) pos-sesses eigenvalues. Their number is countably infinite if and only if thekernel is not degenerate. All eigenvalues of a real symmetric kernel arereal.

Proof: First, recall the Schwarz inequality, in Dirac notation:

|〈f |g〉|2 ≤ 〈f |f〉〈g|g〉. (212)

Consider the “quadratic integral form”:

J(φ, φ) = 〈φ|K|φ〉 ≡∫ b

a

∫ b

ak(x, y)φ(x)φ(y)dxdy, (213)

where φ is any (piecewise) continuous function in [a, b]. We’ll assume |a|, |b| <∞ for simplicity here; the reader may consider what additional criteria mustbe satisifed if the interval is infinite.

Our quadratic integral form is analogous with the quadratic form forsystems of linear equations:

A(x, x) =n∑

i,j=1

aijxixj = ( x )

(A

)(x), (214)

and this analogy persists in much of the discussion, lending an intuitiveperspective.

Notice that if we write:

J(φ, φ) = 〈u|v〉 =∫ ∫

u(x, y)v(x, y)dxdy, (215)

where

u(x, y) ≡ k(x, y) (216)

v(x, y) ≡ φ(x)φ(y), (217)

we have defined a scalar product between the vectors u and v. We are thusled to consider its square,

[J(φ, φ)]2 =∫dx∫dy∫dx′

∫dy′k(x, y)φ(x)φ(y)k(x′, y′)φ(x′)φ(y′), (218)

to which we apply the Schwarz inequality:

[J(φ, φ)]2 = |〈u|v〉|2 ≤ 〈u|u〉〈v|v〉

≤∫ ∫

[φ(x)φ(y)]2 dxdy∫ ∫

[k(x, y)]2 dxdy. (219)

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Thus, if we require φ to be a normalized function,∫[φ(x)]2 dx = 1, (220)

we see that |J(φ, φ)| is bounded, since the integral of the squared kernel isbounded.

Furthermore, we can have J(φ, φ) = 0 for all φ if and only if k(x, y) = 0.The “if” part is obviously true; let us deal with the “only if” part. Thisstatement depends on the symmetry of the kernel. Consider the “bilinearintegral form”:

J(φ, ψ) = J(ψ, φ) ≡∫ ∫

k(x, y)φ(x)ψ(y)dxdy. (221)

We haveJ(φ+ ψ, φ+ ψ) = J(φ, φ) + J(ψ, ψ) + 2J(φ, ψ), (222)

for all φ, ψ piecewise continuous on [a, b]. We see that J(φ, φ) = 0 for all φonly if it is also true that J(φ, ψ) = 0, ∀φ, ψ.

In particular, let us take

ψ(y) =∫k(x, y)φ(x)dx. (223)

Then

0 = J(φ, ψ) =∫dx∫dyk(x, y)φ(x)

∫dx′k(x′, y)φ(x′)

=∫ [∫

k(x, y)φ(x)dx]2

dy. (224)

Thus,∫k(x, y)φ(x)dx = 0 ∀φ. In particular, take for any given value of y,

φ(x) = k(x, y). Then ∫[k(x, y)]2 dx = 0, (225)

and we find k(x, y) = 0.We now assume that J(φ, φ) 6= 0. Let us assume for convenience that

J(φ, φ) can take on positive values. If not, we could repeat the followingarguments for the case J(φ, φ) ≤ 0 ∀φ. We are interested in finding thenormalized φ for which J(φ, φ) attains its greatest possible value. SinceJ(φ, φ) is bounded, there exists a least upper bound:

J(φ, φ) ≤ Λ1 = 1/λ1, ∀φ such that 〈φ|φ〉 = 1. (226)

We wish to show that this bound is actually achieved for a suitable φ(x).Let us suppose that the kernel is uniformly approximated by a series of

degenerate symmetric kernels:

An(x, y) =an∑i,j=1

c(n)ij ωi(x)ωj(y), (227)

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where c(n)ij = c

(n)ji and 〈ωi|ωj〉 = δij, and such that the approximating kernels

are uniformly bounded in the senses:∫ b

a[An(x, y)]2dy ≤ MA, (228)∫ b

a

[∂An∂x

(x, y)

]2

dy ≤ M ′A, (229)

where MA and M ′A are finite and independent of n.

We consider the quadratic integral form for the approximating kernels:

Jn(φ, φ) ≡∫ ∫

Anφ(x)φ(y)dxdy

=an∑i,j=1

c(n)ij

∫ωi(x)φ(x)dx

∫ωj(y)φ(y)dy

=an∑i,j=1

c(n)ij uiuj, (230)

where ui ≡∫ωi(x)φ(x)dx. This is a quadratic form in the numbers u1, u2, . . . , uan .

Now, ∫ [φ(x)−

an∑i=1

uiωi(x)

]2

dx ≥ 0, (231)

implies that (Bessel inequality):

〈φ|φ〉 = 1 ≥an∑i=1

u2i . (232)

The maximum of J(φ, φ) is attained when

an∑i=1

u2i = 1. (233)

More intuitively, note that

φ(x) =an∑i=1

uiωi(x), (234)

unless there is a component of φ orthogonal to all of the ωi. By removingthat component we can make Jn(φ, φ) larger.

We wish to find a function φn(x) such that the maximum is attained. Weknow that it must be of the form φn(x) = u1ω1(x) +u2ω2(x) + · · ·uanωan(x),where

∑u2i = 1, since then 〈φn|φn〉 = 1. The problem of finding max [Jn(φ, φ)]

is thus one of finding the maximum of the quadratic form subject to theconstraint

∑u2i = 1. We know that such a maximum exists, because a con-

tinuous function of several variables, restricted to a finite domain, assumes a

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maximum value in the domain. Suppose that {u} is the appropriate vector.Then

an∑i,j=1

c(n)ij uiuj = Λ1n (235)

is the maximum value that Jn(φ, φ) attains.But the problem of finding the maximum of the quadratic form is just the

problem of finding its maximum eigenvalue and corresponding eigenvector.That is,

an∑j=1

c(n)ij uj = Λ1nui, i = 1, 2, . . . , an. (236)

This is also called the “principal axis problem”. Take

φn(x) = u1ω1(x) + u2ω2(x) + · · ·+ uanωan(x), (237)

where {u} is now our (normalized) vector for which the quadratic form ismaximal. The normalization 〈φn|φn〉 still holds. Apply the approximatekernel operator to this function:∫

An(x, y)φn(y)dy =an∑i,j=1

c(n)ij ωi(x)

∫ωj(y)φn(y)dy

=an∑i=1

ωi(x)an∑j=1

c(n)ij uj

= Λ1n

an∑i=1

uiωi(x)

= Λ1nφn(x). (238)

Therefore φn(x) is an eigenfunction of An(x, y) belonging to eigenvalue λ1n =1/Λ1n.

Finally, it is left to argue that, as we let An converge on k, φn(x) convergeson eigenfunction φ(x), with eigenvalue λ1. We’ll let n → ∞. Since An(x, y)is uniformly convergent on k(x, y), we have that, given any ε > 0, there existsan N such that whenever n ≥ N :

|k(x, y)− An(x, y)| < ε, ∀x, y ∈ [a, b]. (239)

Thus,

[J(φ, φ)− Jn(φ, φ)]2 ={∫ ∫

[k(x, y)− An(x, y)]φ(x)φ(y)dxdy}2

≤ |〈φ|φ〉|2∫ ∫

[k(x, y)− An(x, y)]2 dxdy (Schwarz),

≤ ε2∫ b

a

∫ b

adxdy

≤ ε2(b− a)2. (240)

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Thus, the range of Jn may be made arbitrarily close to the range of J by tak-ing n large enough, and hence, the maximum of Jn may be made arbitrarilyclose to that of J :

limn→∞

Λ1n = λ1. (241)

Now, by the Schwarz inequality, the functions φn(x) are uniformly boundedfor all n:

[φn(x)]2 =[λ1n

∫An(x, y)φn(y)dy

]2

≤ λ21n〈φn|φn〉

∫[An(x, y)]2 dy. (242)

As n → ∞, λ1n → λ1 and An(x, y) → k(x, y). Also, since An(x, y) ispiecewise continuous, φn(x) is continuous, since it is an integral function.The φn(x) form what is known as an “equicontinuous set”: For every ε > 0,there exists δ(ε) > 0, independent of n, such that

|φn(x+ η)− φn(x)| < ε, (243)

whenever |η| < δ. This may be seen as follows: First, we show that φ′n(x) isuniformly bounded:

[φ′n(x)]2

=

[λ1n

∫ ∂An∂x

(x, y)φn(y)dy

]2

≤ λ21n

∫ [∂An∂x

(x, y)

]2

dy (Schwarz)

≤ λ21nM

′A. (244)

Or, [φ′n(x)]2 ≤M ′′A, where M ′′

A = M ′A maxλ2

1n. With this, we find:

|φn(x+ η)− φn(x)|2 =∣∣∣∣∫ x+η

xφ′n(y)dy

∣∣∣∣2=

∣∣∣∣∣∫ b

a[θ(y − x)− θ(y − x− η)]φ′n(y)dy

∣∣∣∣∣2

≤∫ b

a[θ(y − x)− θ(y − x− η)]2 dy

∫ b

a[φ′n(y)]

2dy

≤ |η|(b− a)M ′′A

< ε, (245)

for δ ≤ ε/(b− a)M ′′A.

For such sets of functions there is a theorem analogous to the Bolzano-Weierstrass theorem on the existence of a limit point for a bounded infinitesequence of numbers:

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Theorem: (Arzela) If f1(x), f2(x), . . . is a uniformly bounded equicontin-uous set of functions on a domain D, then it is possible to select asubsequence that converges uniformly to a continuous limit function inthe domain D.

The proof of this is similar to the proof of the Bolzano-Weierstrass theorem,which it relies on. We start by selecting a set of points x1, x2, . . . that iseverywhere dense in [a, b]. For example, we could pick successive midpointsof intervals. By the Bolzano-Weierstrass theorem, this sequence of num-bers contains a convergent subsequence. Now select an infinite sequence offunctions (out of {f}) a1(x), a2(x), . . . whose values at x1 form a convergentsequence, which we may also accomplish by the same reasoning. Similarly,select a convergent sequence of functions (out of {a}) b1(x), b2(x), . . . whosevalues at x2 form a convergent sequence, and so on.

Now consider the “diagonal sequence”:

q1(x) = a1(x)

q2(x) = b2(x)

q3(x) = c3(x)

. . . (246)

We wish to show that the sequence {q} converges on the entire interval [a, b].Given ε > 0, take M large enough so that there exist values xk with

k ≤M such that |x− xk| ≤ δ(ε) for every point x of the interval, where δ(ε)is the δ in our definition of equicontinuity. Now choose N = N(ε) so that form,n > N

|qm(xk)− qn(xk)| < ε, k = 1, 2, . . . ,M. (247)

By equicontinuity, we have, for some k ≤M :

|qm(x)− qm(xk)| < ε, (248)

|qn(x)− qn(xk)| < ε. (249)

Thus, for m,n > N :

|qm(x)− qn(x)| = |qm(x)− qm(xk) + qm(xk)− qn(xk) + qn(xk)− qn(x)|< 3ε. (250)

Thus, {q} is uniformly convergent for all x ∈ [a, b].With this theorem, we can find a subsequence φn1 , φn2 , . . . that converges

uniformly to a continuous limit function ψ1(x) for a ≤ x ≤ b. There may bemore than one limit function, but there cannot be an infinite number, as weknow that the number of eigenfunctions for given λ is finite. Passing to thelimit,

〈φn|φn〉 = 1 → 〈ψ1|ψ1〉 = 1 (251)

Jn(φn, φn) = Λ1n → J(ψ1, ψ1) = Λ1 (252)

φn(x) = λ1n

∫An(x, y)φn(y)dy → ψ1(x) = λ1

∫k(x, y)ψ1(y)dy.(253)

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Thus we have proven the existence of an eigenvalue (λ1).Note that λ1 6=∞ since we assumed that J(φ, φ) could be positive:

max [J(φ, φ)] = Λ1 =1

λ1

> 0. (254)

Note also that, just as in the principal axis problem, additional eigenvalues (ifany exist) can be found by repeating the procedure, restricting to functionsorthogonal to the first one. If k(x, y) is degenerate, there can only be a finitenumber of them, as the reader may demonstrate. This completes the proofof the theorem stated at the beginning of the section.

We’ll conclude this section with some further properties of symmetric ker-nels. Suppose that we have found all of the positive and negative eigenvaluesand ordered them by absolute value:

|λ1| ≤ |λ2| ≤ . . . (255)

Denote the corresponding eigenfunctions by

β1, β2, . . . (256)

They form an orthonormal set (e.g., if two independent eigenfunctions cor-responding to the same eigenvalue are not orthogonal, we use the Gram-Schmidt procedure to obtain orthogonal functions).

We now note that if there are only a finite number of eigenvalues, thenthe kernel k(x, y) must be degenerate:

k(x, y) =n∑i=1

βi(x)βi(y)

λi. (257)

We may demonstrate this as follows: Consider the kernel

k′(x, y) = k(x, y)−n∑i=1

βi(x)βi(y)

λi, (258)

and its integral form

J ′(ψ, ψ) =∫ ∫

k′(x, y)ψ(x)ψ(y)dxdy. (259)

The maximum (and minimum) of this form is zero, since the eigenvalues of

k(x, y) equal eigenvalues of∑ni=1

βi(x)βi(y)λi

. Hence k′(x, y) = 0.We also have the following “expansion theorem” for integral transforms

with a symmetric kernel.

Theorem: Every continuous function g(x) that is an integral transform withsymmetric kernel k(x, y) of a piecewise continuous function f(y),

g(x) =∫k(x, y)f(y)dy, (260)

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where k(y, x) = k(x, y), can be expanded in a uniformly and absolutelyconvergent series in the eigenfunctions of k(x, y):

g(x) =∞∑i=1

giβi(x), (261)

where gi = 〈βi|g〉.

We notice that for series of the form:

k(x, y) =∞∑i=1

βi(x)βi(y)

λi(262)

the theorem is plausible, since

g(x) =∞∑i=1

βi(x)

λi

∫βi(y)f(y)dy

=∞∑i=1

giβi(x), (263)

where gi = 〈βi|f〉/λi = 〈βi|g〉, and we should properly justify the interchangeof the summation and the integral. We’ll forego a proper proof of the theoremand consider its application.

We wish to solve the inhomogeneous integral equation:

g(x) = f(x)− λ∫ b

ak(x, y)f(y)dy. (264)

Suppose that λ is not an eigenvalue, λ 6= λi, i = 1, 2, . . .. Write

f(x)− g(x) = λ∫ b

ak(x, y)f(y)dy. (265)

Assuming f(y) is at least piecewise continuous (hence, f − g must be con-tinuous), the expansion theorem tells us that f(x)− g(x) may be expandedin the absolutely convergent series:

f(x)− g(x) =∞∑i=1

aiβi(x), (266)

where

ai = 〈βi|f − g〉

= λ∫ ∫

k(x, y)f(y)βi(x)dydx

= λ∫f(y)dy

∫k(y, x)βi(x)dx

λi〈βi|f〉. (267)

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Using the first and final lines, we may eliminate 〈βi|f〉:

〈βi|f〉 =λi

λi − λ〈βi|g〉, (268)

and arrive at the result for the expansion coefficients:

ai =λ

λi − λ〈βi|g〉. (269)

Thus, we have the solution to the integral equation:

f(x) = g(x) + λ∞∑i=1

βi(x)〈βi|g〉λi − λ

. (270)

This solution fails only if λ = λi is an eigenvalue, except that it remains valideven in this case if g(x) is orthogonal to all eigenfunctions corresponding toλi, in which case any linear combination of such eigenfunctions may be addedto solution f .

6.1 Resolvent Kernels and Formal Solutions

In the context of the preceding discussion, we may define a “resolvent kernel”R(x, y;λ) by:

f(x) = g(x) + λ∫ b

aR(x, y;λ)g(y)dy. (271)

Then

R(x, y;λ) =∞∑i=1

βi(y)βi(x)

λi − λ. (272)

The Fredholm series:1

λ

D(x, y;λ)

D(λ)(273)

is an example of such a resolvent kernel.Now look at the problem formally: We wish to solve the (operator) equa-

tion f = g + λKf . The solution in terms of the resolvent is

f = g + λRg = (1 + λR)g. (274)

But we could have also obtained the “formal” solution:

f =1

1− λKg. (275)

If “|λK|”< 1 then we have the series solution:

f = g + λKg + λ2K2g + . . . , (276)

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which is just the Neumann series.What do these formal operator equations mean? Well, they only have

meaning in the context of operating on the appropriate operands. For exam-ple, consider the meaning of |λK| < 1. This might mean that for all possiblenormalized functions φ we must have that ‖λK‖ < 1, where the ‖ indicatesan “operator norm”, given by:

‖λK‖ ≡ maxφ

[λ∫ ∫

k(x, y)φ(x)φ(y)dxdy]2

< 1. (277)

By the Schwarz inequality, we have that

‖λK‖ ≡ λ2∫ ∫

[k(x, y)]2 dxdy. (278)

The reader is invited to compare this notion with the condition for con-vergence of the Neumann series in Whittaker and Watson:

|λ(b− a)|maxx,y|k(x, y)| < 1. (279)

6.2 Example

Consider the problem:

f(x) = sin2 x+ λ∫ 2π

0k(x, y)f(y)dy, (280)

with symmetric kernel

k(x, y) =1

1− α2

1− 2α cos(x− y) + α2, |α| < 1. (281)

We look for a solution of the form

f(x) = g(x) + λ∞∑i=1

〈βi|g〉λi − λ

βi(x), (282)

where g(x) = sin2 x. In order to accomplish this, we need to determine theeigenfunctions of the kernel.

With some inspection, we realize that the constant 1/√

2π is an (normal-ized) eigenfunction. This is because the integral:

I0 ≡∫ 2π

0

dx

1− 2α cos(x− y) + α2(283)

is simply a constant, with no dependence on y. In order to find the cor-responding eigenvalue, we must evaluate I0. Since I0 is independent of y,evaluate it at y = 0:

I0 =∫ 2π

0

dx

1− 2α cosx+ α2. (284)

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We turn this into a contour integral on the unit circle, letting z = eix.Then dx = dz/iz and 2 cosx = z + 1

z. This leads to:

I0 = i∮ dz

αz2 − (1 + α2)z + α. (285)

The roots of the quadratic in the denominator are at z = {α, 1/α}. Thus,

I0 =i

α

∮ dz

(z − α)(z − 1/α). (286)

Only the root at α is inside the contour; we evaluate the residue at this pole,and hence determine that

I0 =2π

1− α2. (287)

We conclude that eigenfunction 1/√

2π corresponds to eigenvalue 1.We wish to find the rest of the eigenfunctions. Note that if we had not

taken y = 0 in evaluating I0, we would have written:

I0 =ieiy

α

∮ dz

(z − eiyα)(z − eiy/α), (288)

and the relevant pole is at eiyα. We thence notice that we know a whole classof integrals:

1− α2

ieiy

α

∮ zndz

(z − eiyα)(z − eiy/α)= αneiny, n ≥ 0. (289)

Since zn = einx, we have found an infinite set of eigenfunctions, and theiregienvalues. But we should investigate the negative powers as well – we didn’tinclude them here so far because they yield an additional pole, at z = 0.

We wish to evaluate:

I−n ≡ieiy

α

∮ dz

zn(z − eiyα)(z − eiy/α), n ≥ 0. (290)

The residue at pole z = αeiy is 11−α2

1αneiny

. We need also the residue at z = 0.It is coefficient A−1 in the expansion:

eiy

zn(z − eiyα)(z − eiy/α)=

∞∑j=−∞

Ajzj. (291)

After some algebra, we find that

eiy

zn(z − eiyα)(z − eiy/α)=

α

1− α2

∞∑j=0

zj−ne−i(j+1)y[

1

αj+1− αj+1

]. (292)

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The j = n− 1 term will give us the residue at z = 0:

A−1 = e−inyα

1− α2

(α−n − αn

). (293)

Thus,

I−n =2π

1− α2αne−iny. (294)

We summarize the result: The normalized eigenfunctions are βn(x) =einx√

2π, with eigenvalues λn = α−|n|, for n = 0,±1,±2, . . ..Finally, it remains to calculate:

f(x) = sin2 x+ λ∞∑n=1

〈βn| sin2 x〉λn − λ

βn(x)

= sin2 x+ λ

√2π

4

[2β0(x)

1− λ− β−2(x) + β2(x)

α−2 − λ

]

= sin2 x+λ

2

(1

1− λ− 1

α−2 − λcos 2x

). (295)

Note that if λ = 1 or λ = α−2 then there is no solution. On the other hand,if λ = α−|n| is one of the other eigenvalues (n 6= 0,±2), then the above is stilla solution, but it is not unique, since we can add any linear combination ofβn(x) and β−n(x) and still have a solution.

7 Exercises

1. Given an abstract complex vector space (linear space), upon which wehave defined a scalar product (inner product):

〈a|b〉 (296)

between any two vectors a and b, prove the Schwarz inequality:

|〈a|b〉|2 ≤ 〈a|a〉〈b|b〉. (297)

Give the condition for equality to hold.

One way to approach the proof is to consider the fact that the projectionof a onto the subspace which is orthogonal to b cannot have a negativelength, where we define the length (norm) of a vector according to:

‖c‖ ≡√〈c|c〉. (298)

Further, prove the triangle inequality:

‖a+ b‖ ≤ ‖a‖+ ‖b‖. (299)

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2. Considering our RC circuit example, derive the results in Eqn. 31through Eqn. 35 using the Fourier transform.

3. Prove the convolution theorem.

4. We showed the the Fourier transform of a Gaussian was also a Gaussianshape. That is, let us denote a Gaussian of mean µ and standarddeviation σ by:

N(x;µ, σ) =1√2πσ

exp

[−(x− µ)2

2σ2

]. (300)

(a) In class we found (in an equivalent form) that the Fourier Trans-form of a Gaussian of mean zero was:

N(y; 0, σ) =1√2π

exp

[−y

2σ2

2

]. (301)

Generalize this result to find the Fourier transform of N(x;µ, σ).

(b) The experimental resolution function of many measurements isapproximately Gaussian in shape (in probability&statistics we’llprove the “Central Limit Theorem”). Often, there is more thanone source of uncertainty contributing to the final result. For ex-ample, we might measure a distance in two independent pieces,with means µ1, µ2 and standard deviations σ1, σ2. The resolu-tion function (sampling distribution) of the final result is then theconvolution of the two pieces:

P (x;µ1, σ1, µ2, σ2) =∫ ∞−∞

N(y;µ1, σ1)N(x− y;µ2, σ2)dy. (302)

Do this integral to find P (x;µ1, σ1, µ2, σ2). Note that it is possibleto do so by straightforward means, though it is a bit tedious.You are asked here to instead use Fourier transforms to (I hope!)obtain the result much more easily.

5. The “Gaussian integral” is:

1√2πσ

∫ ∞−∞

exp

[−(x− µ)2

2σ2

]dx = 1. (303)

Typically, the constants µ and σ2 are real. However, we have encoun-tered situations where they are complex. Determine the domain in(µ, σ2) for which this integral is valid. Try to do a careful and convinc-ing demonstration of your answer.

6. In class we consider the three-dimensional Fourier transform of e−µr/r,where r =

√x2 + y2 + z2. What would the Fourier transform be in two

dimensions (i.e., in a two-dimensional space with r =√x2 + y2)?

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7. The lowest P -wave hydrogen wave function in position space may bewritten:

ψ(x) =1√

32πa50

r cos θ exp(− r

2a0

), (304)

where r =√x2 + y2 + z2, θ is the polar angle with respect to the z

axis, and a0 is a constant. Find the momentum-space wave functionfor this state (i.e., find the Fourier transform of this function).

In this and all problems in this course, I urge you to avoid look-uptables (e.g., of integrals). If you do feel the need to resort to tables,however, be sure to state your source.

8. In section 2.2.1, we applied the Laplace transform method to determinethe response of the RC circuit:

R

R C

V(t)

V

1

2

c (t)

to an input voltage V (t) which was a delta function. Now determineVC(t) for a pulse input. Model the pulse as the difference between twoexponentials:

V (t) = A(e−t/τ1 − e−t/τ2

). (305)

9. In considering the homogeneous integral equation, we stated the the-orem that there are a finite number of eigenfunctions for any giveneigenvalue. We proved this for real functions; now generalize the proofto complex functions.

10. Give a graphical proof that the series D(λ) and D(x, y;λ) in the Fred-holm solution are polynomials of degree n if the kernel is of the degen-erate form:

k(x, y) =n∑i=1

φi(x)ψi(y). (306)

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11. Solve the following equation for u(t):

d2u

dt2(t) +

∫ 1

0sin [k(s− t)]u(s)ds = a(t), (307)

with boundary condition u(0) = u′(0) = 0, and a(t) is a given function.

12. Prove that an n-term degenerate kernel possesses at most n distincteigenvalues.

13. Solve the integral equation:

f(x) = ex +∫ x

1

1 + y

xf(y)dy. (308)

Hint: If you need help solving a differential equation, have a look atMathews and Walker chapter 1.

14. In section 5.2.1 we developed an algorithm for the numerical solutionof Volterra’s equation. Apply this method to the equation:

f(x) = x+∫ x

0e−xyf(y)dy. (309)

In particular, estimate f(1), using one, two, and three intervals (i.e.,N = 1, N = 2, and N = 3). [We’re only doing some low values so youdon’t have to develop a lot of technology to do the computation, butgoing to high enough N to get a glimpse at the convergence.]

15. Another method we discussed in section 3 is the extension to theLaplace transform in Laplace’s method for solving differential equa-tions. I’ll summarize here: We are given a differential equation of theform:

n∑k=0

(ak + bkx)f (k)(x) = 0 (310)

We assume a solution of the form:

f(x) =∫CF (s)esxds, (311)

where C is chosen depending on the problem. Letting

U(s) =n∑k=0

aksk (312)

V (s) =n∑k=0

bksk, (313)

the formal solution for F (s) is:

F (s) =A

V (s)exp

∫ s U(s′)

V (s′)ds′, (314)

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where A is an arbitrary constant.

A differential equation that arises in the study of the hydrogen atomis the Laguerre equation:

xf ′′(x) + (1− x)f ′(x) + λf(x) = 0. (315)

Let us attack the solution to this equation using Laplace’s method.

(a) Find F (s) for this differential equation.

(b) Suppose that λ = n = 0, 1, 2, . . . Pick an appropriate contour, anddetermine fn(x).

16. Write the diagram, with coefficients, for the fifth-order numerator anddenominator of the Fredholm expansion.

17. Solve the equation:

f(x) = sinx+ λ∫ π

0cosx sin yf(y)dy (316)

for f(x). Find any eigenvalues and the corresponding eigenfunctions.Hint: This problem is trivial!

18. Find the eigenvalues and eigenfunctions of the kernel:

k(x, y) =1

2log

∣∣∣∣sin x+ y

2/ sin

x− y2

∣∣∣∣=

∞∑n=1

sinnx sinny

n, 0 ≤ x, y ≤ π. (317)

19. In the notes we considered the kernel:

k(x, y) =1

1− α2

1− 2α cos(x− y) + α2, (318)

where |α| < 1 and 0 ≤ x, y ≤ 2π. Solve the integral equation

f(x) = ex + λ∫ 2π

0k(x, y)f(y)dy (319)

with this kernel. What happens if λ is an eigenvalue? If your solutionis in the form of a series, does it converge?

20. Solve for f(x):

f(x) = x+∫ x

0(y − x)f(y)dy, (320)

This problem can be done in various ways. If you happen to obtain aseries solution, be sure to sum the series.

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21. We wish to solve the following integral equation for f(x):

f(x) = g(x)− λ∫ x

0f(y)dy, (321)

where g(x) is a known, real continuous function with continuous firstderivative, and satisfies g(0) = 0.

(a) Show that this problem may be re-expressed as a differential equa-tion with suitable boundary condition, which may be written inoperator form as Lf = g′. Give L explicitly and show that it is alinear operator.

(b) Suppose that G(x, y) is the solution of LG = δ(x − y), whereδ(x) is the Dirac δ function. Express the solution to the originalproblem in the form of an integral transform involving G and g′.

(c) Find G(x, y) and write down the solution for f(x).

22. Some more Volterra’s equations: Solve for f(x) in the following twocases –

(a) f(x) = sinx+ cosx+∫ x

0 sin(x− y)f(y) dy,

(b) f(x) = e−x + 2x+∫ x

0 ey−xf(y) dy.

23. Consider the LCR circuit in Fig. 4:

L

R C

V(t)

V (t)0

Figure 4: An LCR circuit.

Use the Laplace transform to determine V0(t) given

V (t) ={V 0 < t < T0 otherwise.

(322)

Make a sketch of V (t) for (a) 2RC >√LC; (b) 2RC <

√LC; (c)

2RC =√LC.

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24. The radioactive decay of a nucleus is a random process in which theprobability of a decay in time interval (t, t+dt) is independent of t, if thedecay has not already occurred. This leads to the familiar exponentialdecay law (as you may wish to convince yourself): If at time t, thereare N(t) nuclei, then the rate of decays is proportional to N(t):

dN

dt(t) = −λN(t).

Integrating, we find N(t):

N(t) = N(0)e−λt.

In practice, radioactive decays often occur in long chains. For (a sim-plified) example, 238U decays via α-emission to 234Th with a half-life of4.5 × 109 y; 234Th decays in a subchain with two β emissions to 234Uwith a half-life of 24 d; 234U decays via α-emission to 230Th with a2.4×106 y half-life; etc. We may use the method of Laplace transformsto determine how the abundance of any species of nucleus in such achain evolves with time.

Thus, suppose that we have a decay chain A → B → C → D, whereD is stable, and the decay rates for A, B, and C are λA, λB, and λC ,respectively. Suppose NB(0) = 0. Determine NC(t) as a function ofthe rates and the initial abundances NA(0) and NC(0).

You are supposed to approach this problem by setting up a systemof differential equations for the abundances, and then using Laplacetransforms to solve the equations. Note, in setting up your differentialequations, that the rate of change in abundance for an intermediatenucleus in the chain gets a contribution from the nucleus decaying intoit as well as from its own decay rate.

25. Solve the following integral equations for f(x):

(a)

f(x) = ex + λ∫ 2

0xyf(y)dy. (323)

(b)

f(x) = λ∫ π

0sin(x− y)f(y)dy. (324)

For both parts, what happens for different values of λ?

26. Consider the following simple integral equation:

f(x) = x2 + λ∫ 1

0xyf(y)dy. (325)

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(a) Find the Neumann series solution, to order λ2. For what valuesof λ do you expect the Neumann series to be convergent? If youaren’t sure from what you have done so far, try doing the rest ofthe problem and come back to this.

(b) Find the Fredholm series solution, to order λ2.

(c) This is a degenerate kernel, so find the solution according to ourmethod for degenerate kernels.

48