physics problem

advanced physical problems involved with  Fourier transform ,  Frobenius procedure,  Laplace transforms, and so on. the answer has to be typeset. 

Problem 1:

A: The function f(x) is the Fourier transform of the function

F (p) =

2

e−ıpL − 1− e−2ıpL√

2π p2
(1)

Use the techniques we developed to calculate integrals in the complex plane to find the
explicit expression of

f(x) =
1√
2π

∫ ∞
−∞

F (p) eıpx dp (2)

B: The function g(x) is the Fourier transform of the function

G(p) = F (p)

1

1− e−2ıpL−�
=

2e−ıpL − 1− e−2ıpL√
2π p2

1

1− e−2ıpL−�
(

3

)

where � is a very small positive parameter which can be sent to zero at the end of the
calculations. Find the explicit expression of the function

g(x) =
1√
2π

∫ ∞
−∞

G(p) eıpx dp (

4

)

Hint: Use Eq.

5

1 in Lecture 5.

Problem 2:

The equation
d2y(x)

dx2
+

1

x

dy(x)

dx
+
x2 − 4
x2

y(x) = 0 (5)

has a regular singular point at x = 0.

A: What are the solutions of the indicial equation?

B: Denote the two solutions by s1, s2 with s1 < s2. Use the Frobenius procedure to find the first four non-vanishing coefficients in the power series expansion of the solution y2(x) corresponding to the root s2:

y2(x) = x
s2 + . . . (6)

Note that in this equation we are taking the first coefficient of the expansion to be equal to
1. So you really only need to find the next three non-vanishing coefficients.

C: Show that the Frobenius procedure fails if you try to obtain a power series expansion for
the other solution of the form

y1(x) = x
s1 + . . . (7)

Show in detail when and why it fails.

D: Without doing any actual calculation, explain how you might find the other solution.

2

VB

R0

C

v(t) q(t)

i(t)

t

v(t)

V1

V2

T T ′

Figure 1: Illustration of the circuit described in the introduction and of the periodic signal
it generates.

Problem 3:

Introduction. These introductory paragraphs provide a context to the problem but are not
needed to solve it. You may skip them and proceed to the question A and the following
questions.

A simple periodic signal can be obtained with a battery, a resistor, a capacitor, and a neon
light, as shown in Fig. 1. When the voltage v(t) on the neon light, represented by the
orange circle, is below a certain threshold V2 < VB, the gas inside the bulb offers a very high resistance to the passage of current and the capacitor charges through the resistor R0 causing v(t) to increase. When v(t) reaches the value V2 the gas ionizes and the neon light conducts current, offering very low resistance. The capacitor discharges through the light causing the v(t) to drop. When v(t) reaches the value V1 the voltage on the neon light is no longer sufficient to sustain the ionization, the passage of current through the light essentially stops and the cycle begins again. The circuit with the neon light can be well approximated by the circuit show in Fig. 2, with the resistance R switching from a very high value when v(t) grows from V1 to V2 to a very low value when v(t) decreases from V2 to V1. This motivates the questions in this problem, which are, however, independent of the above considerations.

A: Consider the circuit shown in Fig. 2. See the figure caption for the details of the circuit.
Write the equations that govern the behavior in time of v(t), i.e. the potential at the top
plate of the capacitor (the bottom plate is grounded, as the figure shows), and of the other
components of the circuit. Use the Laplace transform to solve the equations, finding the
behavior of v(t) with time assuming the its initial value is

v(0) = v0 < VB (8)

3

VB
R0
C

R

v(t) q(t)
i(t)

Figure 2: The circuit for question A. The battery provides a voltage VB. The resistor
at the top of the circuit has resistance R0 and i(t) denotes the intensity of the current
flowing through the resistor. The capacitor has capacitance C, q(t) denotes the charge in
the capacitor and v(t) the voltage on its top plate as well as at the top of the resistor with
resistance R, which is connected in parallel to the capacitor. The other plate of the capacitor
as well as the other terminal of the resistor are grounded, as the diagram shows. (Reminders
of electrodynamics: the voltage drop across a resistor is equal to the product of the resistance
times the intensity of the current flowing through it; the charge on a capacitor is equal to
the product of its capacitance times the voltage across the plates.)

Please note: it is possible to solve the equations without using the Laplace transform, but
this is a problem on the Laplace transform, so to get credit for this question you should:

• Write the differential equation(s) that determine the time behavior of v(t).

• Perform the Laplace transform of these equations, which will convert them into alge-
braic equations.

• Solve these equations to find the Laplace transform V (s) of v(t).

• Use the result obtained for V (s) to find the time dependence of v(t).

B: The Laplace transforms F (s) and G(s) of the functions f(t)θ(t) and g(t)θ(t) are given by
the following equations

F (s) =
a(1− e−st)

s
(9)

G(s) =
b(1− e−(s+c)T )

s+ c
(10)

(11)

4

where a, b, c, T are some positive constants. Find the values of f(t) and g(t) for t ≥ 0. (As
a reminder: the function θ(t) is defined as θ(t) = 0 for t < 0, θ(t) = 1 for t ≥ 0.) C: Let the function ṽ(t) be equal to the solution v(t) obtained in A for 0 ≤ t ≤ T and equal to zero for t < 0 and t > T , i.e.

ṽ(t) = v(t)θ(t)θ(T − t) (12)

Find its Laplace transform Ṽ (s).

(You may take advantage of the results in B, or calculate Ṽ (s) from your solution for v(t)
and the definition of the Laplace transform.)

Problem 4:

Consider the function φ(x, y) in the domain D defined by x > 0, y > 0 (i.e. D is the upper
right quadrant of the x− y plane.)
In the domain D the function φ(x, y) satisfies the equation

x2
∂

2φ(x, y)

∂x2
− 2xy∂

2φ(x, y)

∂x ∂y
+my2

∂2φ(x, y)

∂x2
= 0 (13)

or, with concise notation,
x2φxx − 2xyφxy +my2φyy = 0 (14)

where m is a real constant.

A: For what values of m will the equation be an elliptic, parabolic, or hyperbolic equation?

B: Find the characteristics of Eq. 14 with m = −3, i.e. find the characteristics of the equation

x2φxx − 2xyφxy − 3y2φyy = 0 (15)

within the domain D.

5

f = (f0,f1, . . .fN−

2

,fN−1) into f
′ = (f1,f2, . . .fN−1,f0). The transformation f → f ′ = Cf

is linear and its matrix representation has the form:

C =




0 1 0 . . . 0 0
0 0 1 . . . 0 0
. . . . . . . . . . . .

. . . . . .

0 0 0 . . . 1 0
0 0 0 . . . 0

1

1 0 0 . . . 0 0


 (1)

C is unitary, since, obviously, f ′2 = f2. Its inverse C−1 = C† permutes the elements of f in
the opposite order:

C−1 =




0 0 . . . 0 0 1
1 0 . . . 0 0 0
0 1 . . . 0 0 0
. . . . . . . . . . . . . . . . . .
0 0 . . . 1 0 0
0 0 . . . 0 1 0


 (2)

After N circular shifts, the components of the vector f return to the original order. Thus
CN = I. It follows that, given any eigenvalue λ of C, λN = 1: λ must be one of the Nth

roots of unity. The possible eigenvalues of C will therefore be

λk = e
2πık
N , k = 0 . . .N − 1 (

3

)

It is convenient to define
z = e

2πı

N (

4

)

With this notation
λk = z

k, k = 0 . . .N − 1 (

5

)

We will soon see that each of these eigenvalues occurs exactly once.

Starting from the possible eigenvalue λk, we can try solving for the components of the
corresponding eigenvector f(k). The eigenvalue equation

Cf(k) = λkf
(k) = zkf(k) (

6

)

1

These notes cannot be duplicated and distributed without explicit permission of the author.

Eigenvalues and eigenvectors of the circular shift operator and the finite Fourier transform.

Let us consider vectors in a space of dimension N. The circular shift operator maps the vector

Lecture 5. The Fourier transform

gives (cfr. Eq. 1)

f
(k)
1 = z

kf
(k)
0

f
(k)
2 = z

kf
(k)
1

. . . . . .

f
(k)
N−1 = z

kf
(k)
N−

2

f

(k)
0 = z

kf
(k)
N−1 (

7

)

The first N − 1 equations determine

f
(k)
j = z

jkf
(k)
0 (

8

)

The last equation
f

(k)
0 = z

kf
(k)
N−1 = z

kNf
(k)
0 (

9

)

is a consistency condition, which is satisfied because zkN = 1. Thus we see that all λk =
zk, k = 0 . . .N − 1, are indeed eigenvalues of C. Since there are N of them, they exhaust
all the eigenvalues and each occurs exactly once.

We fix the eigenvectors uniquely by

f
(k)
0 =

1
√

N

(

10

)

This gives

f
(k)
j =

1

√
N

e

2πıjk
N (

11

)

for the eigenvector components and the normalization

|f(k)| =
√∑

j

f
(k)∗
j f

(k)
j = 1 (

12

)

Eigenvectors corresponding to different values of k are orthogonal. This follows from the
general properties of eigenvectors of unitary matrices and can also easily be checked explicitly:
for k 6= k′

f(k)f(k
′) =

∑
j

f
(k)∗
j f

(k′)
j =

∑
j

z−kjzk
′j =

∑
j

z(k
′−k)j =

1 −z(k′−k)N

1 −z(k′−k)
= 0 (

13

)

The eigenvectors f(k) thus form an orthonormal basis. Any vector f can be expanded into
this basis:

f =

∑
k

Fkf
(k) (

14

)

2

Explicitly

fj =
∑
k

Fkf
(k)
j =

1
√
N
∑
k

Fk e
2πıjk
N (

15

)

Since this is an orthonormal change of basis, the norm of the vector is preserved∑
j

|fj|2 =
∑
k

|Fk|2 (

16

)

The change of basis expressed by Eqs. 14, 15 goes under the name of (finite) Fourier trans-
form. The numbers Fk, k = 0 . . .N − 1 are the Fourier components of f.

Properties of the finite Fourier transform

The Fourier components of f can be found by using the orthonormality of the eigenvectors:

Fk = f
∗(k)f =

1
√
N
∑
j

fj e
−2πıjk
N (17)

The action of the circular shift operator C takes a particularly simple form when expressed
in terms of the Fourier components. From the fact that C is diagonal in the basis of its
eigenvectors it follows that, if f ′ = Cf,

F ′k = λkFk = e
2πık
N Fk (18)

This can also be seen directly from the expansion 14

f ′ = Cf =
∑
k

FkCf
(k) =

∑
k

λkFkf
(k) (19)

which shows that F ′k = λkFk, as in Eq. 18 above. The fact that C is diagonal in Fourier
space is especially relevant because C enters in most numerical approximations to differential
operators. For example, the central difference approximation to the second derivative is the
operator

D2 =
C + C† − 2I

a2
(20)

where we assumed that the numbers fj represent the value of some function f(x) at the
points xj = x0 + ja. In Fourier space, D2 is also diagonal

(D2F)k =
e

2πık
N + e

−2πık
N − 2

a2

Fk =

2 cos
(

2πk
N

)
− 2

a2
Fk (21)

For several considerations that follow, it will be convenient to extend the range of indices
j,k etc. beyond 0,N − 1. We will assume that all indices are defined mod N, so that

fj+mN = fj, Fk+mN = Fk (22)

3

with integer m. This convention is consistent with the relation between fj and Fk (Eqs. 15, 17)
since exp[2πı(j + mN)k/N] = exp[2πıj(k + mN)/N] = exp[2πıjk/N].

Given three vectors f,f ′,f ′′ we say that f ′′ is the convolution of f and f ′ if

f ′′i =
1
√
N

N−1∑
j=0

fi−jf
′
j (23)

where we use the convention just introduced above for the case when the index i− j takes
negative values. The vector with components fi−j can be obtained from f by the action of
C−j. Thus

f ′′ =
1
√
N

∑
j

f ′j C
−jf (24)

In Fourier space this becomes

F ′′k =
1
√
N

∑
j

f ′j e
−2πıjk
N Fk (25)

which, on account of Eq. 17, gives
F ′′k = FkF

′
k (26)

This is the “convolution theorem” for Fourier transforms. It shows that the convolution
operation becomes just a component-by-component product in Fourier space.

In some applications the vector f will have real components. Its Fourier components Fk, in
general, will still be complex. The implication of having real fj can be derived by taking the
complex conjugate of Eq. 17

F∗k =
∑
j

f∗j e
2πıjk
N =

∑
j

fje
−2πıj(−k)

N = F−k (27)

Thus the reality of f implies the constraints F∗k = F−k on the its Fourier components. These
constraints are equivalent to N equations on the 2N real variables ReFk, ImFk, so the total
number of independent real variables is preserved.

The Fourier series

The Fourier transform can be extended to functions of a continuous variable. We will
proceed in two steps. We will consider first functions f(x) defined over the finite range
−L/2 ≤ x < L/2 and will later extend the range to the whole real axis by letting L →∞. Let us divide the interval −L/2 ≤ x < L/2 into an even number N of subintervals of width a = L/N. (For convenience, we will take N to be even, although this is not crucial.) We approximate the function f(x) by the values fj it takes at the points xj = ja:

fj = f(ja), j = −
N

2
, . . .

N

2
− 1 (28)

4

We take the Fourier transform of f, but allow for a different normalization

Fk =
α
√
N

∑
j

fj e
−2πıjk
N (29)

where α is a normalization factor which will be specified below. (Remember that indices can
be thought of as defined mod N, so that, although we are letting now j vary from −N/2 to
(N/2)−1, this formula only differs from Eq. 17 in the normalization.) We rewrite Eq. 29 in
terms of x and f(x)

Fk =
α
√
N
∑
j

f(xj) e
−2πıxjk

L (30)

We set now α =
√
a. With this normalization, Eq. 30 becomes

Fk =

√
a

√
N
∑
j
f(xj) e
−2πıxjk

L =

1
√

L

∑
j

af(xj) e
−2πıxjk

L (31)

We recognize now in the r.h.s. of this equation the numerical approximation to the integral
of f(x) exp(−2πıkx/L). We take the continuum limit by letting N →∞, a → 0 and obtain

Fk =
1
√
L

∫ L/2
−L/2

f(x) e
−2πıkx

L dx (32)

This is the Fourier transform for functions defined over a finite range. It maps f(x) into an
infinite set of Fourier coefficients Fk, k = −∞, . . .∞. The transformation preserves the
norm, in the sense that ∫ L/2

−L/2
|f(x)|2 dx =

∑
k

|Fk|2 (33)

This follows from ∑
k

|Fk|2 = a
∑
j

|fj|2 (34)

(cfr. Eq. 16) and motivated our choice of normalization. The inverse of the transformation 32
can be obtained from

fj =
1
√
aN

∑
k

Fk e
2πıjk
N =

1
√
L
∑
k

Fk e
2πıxjk

L (35)

(cfr. Eq. 15 and remember the change of normalization). In the limit N →∞, a → 0 this
becomes

f(x) =

1

√

L

∑
k

Fk e
2πıkx
L (36)

This constitutes the Fourier series representation of a function defined over a finite range
with periodic boundary conditions. Other Fourier series representations of functions defined

5

over a finite interval, like the Fourier sine series or the Fourier cosine series, can be derived
from this one

Notice that Eq. 36 defines a function f(x) for all values of x, not necessarily restricted to
the interval −L/2 ≤ x < L/2. Indeed, f(x), as defined by Eq. 36, is a periodic function of x with period L:

f(x + mL) = f(x) (37)

with integer m. On account of such periodicity, we will drop explicit mention of the range
in all integrals involving f(x), assuming that the integral is over one full period of f.

The continuum equivalent of the circular shift operator is the infinitesimal displacement
f(x) → f(x + dx). The transformation of f under such infinitesimal transformation is
captured by the derivative and it is therefore no surprise that the derivative operator becomes
diagonal in Fourier space. The Fourier components of f ′(x) = df(x)/dx are

F ′k =
2πık

L
Fk (38)

The finite shift f(x) → f̃(x) = f(x + b) is also diagonal in Fourier space, with

F̃k = e
2πıkb
L Fk (39)

and the convolution theorem becomes the following. If f(x), g(x) and h(x) are related by

h(x) =
1
√

L

∫
f(x−y) g(y) dy (40)

then their Fourier components satisfy

Hk = FkGk (41)

Notice that the Fourier transformation defined by Eqs. 32, 36 goes both ways, in the sense
that, as it associates to a function f(x) the infinite set of Fourier coefficients Fk, so, given an
infinite sequence Fk, it can be used to associate with it the periodic function f(x). Sometime
relations involving sequences can thus be converted into simpler relations on the functions
that are their Fourier transforms. For example, the action of shift operator on the sequence,
Fk → F ′k = Fk+1, becomes diagonal on f(x): f(x) → f

′(x) = exp(−2πıx/L)f(x). There is
also the equivalent of the convolution theorem: if Fk,F

′
k and F

′′
k are related by

F ′′k =
1
√
L

∑
k′

Fk−k′ F
′
k′ (42)

then the associated functions f(x),f ′(x) and f ′′(x) are related by

f ′′(x) = f(x)f ′(x) (43)

6

The Fourier integral

Finally, we can let x vary over an infinite range by taking the limit L → ∞. In order to
preserve a meaningful exponent in Eq. 36 we introduce a new variable p related to k by

pk =

2π

L
k (44)

We also allow again for a change of normalization. We thus rewrite Eq. 36 as

f(x) =

β
√
L

∑
k

F(pk) e
ıpkx (45)

We notice that the spacing between subsequent values pk and pk+1 of p is ∆p = 2π/L.
This spacing goes to zero when L goes to infinity and thus, by choosing β so that β/

√
L is

proportional to ∆p, it becomes possible to interpret the r.h.s. of Eq. 45 as the approximation
to an integral. It is convenient to set β =

√
2π/L. Equation 45 becomes then

f(x) =
1
√
2π
∑
k

F(pk) e
ıpkx ∆p (46)

In the limit L →∞, ∆p → 0 the r.h.s. becomes an integral and the limit of Eq. 46 gives the
Fourier transform over a continuum infinite range:

f(x) =
1
√
2π

∫ ∞
−∞

F(p) eıpx dp (47)

The inverse transform can be obtained from Eq. 32 with the appropriate change of variables
and normalization

F(pk) = lim
L→∞

1
β
√
L
∫ L/2
−L/2

f(x) e−ıxpk dx (48)

i.e.

F(p) =
1
√

2π
∫ ∞
−∞

f(x) e−ıpx dx (49)

Notice the symmetry between Eqs. 47 and 49. Our particular choice for the factor β in
Eq. 45 was motivated by the desire of getting the same normalization factor 1/

√
2π in

Eqs. 47 and 49.

The infinite-range, continuous Fourier transform has many useful properties, similar to those
encountered with the finite-range transform. It preserves the norm:∫

|F(p)|2 dp =
∫
|f(x)|2 dx (50)

The derivative and shift operators are diagonal in Fourier space:

g(x) = f ′(x) ⇒ G(p) = ıpF(p)
g(x) = f(x + b) ⇒ G(p) = eıpbF(p) (51)

7

The convolution theorem holds:

h(x) =
1
√
2π

∫
f(x−y)g(y) dy ⇒ H(p) = F(p)G(p) (52)

To conclude, by performing the Fourier transform followed by its inverse, we get

f(x) =
1
2π

∫ [∫
f(y) e−ıpy dy

]
eıpx dp =

∫ [ 1
2π

∫
eıp(x−y) dp

]
f(y) dy (53)

This implies
1

2π
∫ ∞
−∞

eıp(x−y) dp = δ(x−y) (54)

This equation, and its finite-range and discrete counterparts

1
L

∞∑
k=−∞

e
2πık(x−y)

L =
∞∑

m=−∞

δ(x−y −mL) (55)

1
L
∫ L/2
−L/2

e
2πı(k−k′)x

L = δk,k′ (56)

1
N

N−1∑
k=0

e
2πık(j−j′)

N =
∞∑

m=−∞

δj,j′+mN (57)

express the completeness of Fourier space and are quite useful to remember.

A note about quantum mechanics

The Fourier integral plays a crucial role in quantum mechanics where it serves to go from
the configuration space representation of the wave function ψ(x) to its momentum space
representation φ(p). In this context it is useful to deal with a Fourier variable p which has
dimensions of momentum rather then of inverse length, as in the previous section. (The
dimension of p in Eqs. 47 and 49 follows from the fact that the argument ıpx of the expo-
nential must be dimensionless.) For the purpose one defines the Fourier transform, or, more
precisely, the transformations between configuration space and momentum space, with an h̄
at denominator in the exponentials:

ψ(x) =
1
√

2πh̄

∫ ∞
−∞

φ(p) eıpx/h̄ dp (58)

and

φ(p) =
1
√

2πh̄
∫ ∞
−∞

ψ(x) e−ıpx/h̄ dx (59)

The counterpart of Eq. 54 is

1
2πh̄
∫ ∞
−∞

eıp(x−y)/h̄ dp = δ(x−y) (60)

8

A problem illustrating the notion of Green function, the use of the Fourier series, and of
contour integrals in the complex plane.

Consider the differential equation

−
d2f(x)

dx2

= b(x) (61)

where f(x), the function to be determined, and b(x), the known function, are defined over
0 ≤ x ≤ L with Dirichlet boundary conditions, i.e. f(0) = f(L) = 0, b(0) = b(L) = 0.
Since the equation is linear, its solutions f1(x) and f2(x) corresponding to b1(x), b2(x) can
be added to obtain the solution corresponding to b1(x) + b2(x):

−
d2[f1(x) + f2(x)]

dx2
= b1(x) + b2(x) (62)

This generalizes to the superposition of any number of solutions and also to a continuum of
solutions. It follows that if we can find the function G(x,y) which satisfies the equation

−

d2G(x,y)

dx2
= δ(x−y) (63)

the solution for any b(x) can be expressed as an integral involving G(x,y) as follows

f(x) =

∫ L
0

G(x,y) b(y) dy (64)

Indeed, by differentiating we find

−
d2f(x)

dx2
= −

∫ L
0
d2G(x,y)

dx2
b(y) dy =

∫ L
0

δ(x−y)b(y) dy = b(x) (65)

The function G(x,y) is called the Green’s function for the differential equation we are con-
sidering. Please note that, as with all differential equations, the boundary conditions must
be included in the definition of the differential equation.

In the case at hand it is easy to find the Green’s function. Indeed for x 6= y Eq. 63 reduces
to the equation

−
d2G(x,y)

dx2
= 0 (66)

whose solution is a linear function. We must allow, however, for different linear functions in
the ranges 0 ≤ x < y and y < x ≤ L. Thus, mindful of the Dirichlet boundary conditions, we set

G(x,y) = αx for x < y (67)

G(x,y) = β(L−x) for x > y (68)

9

Figure 1: A colorful 3D representation of the Green’s function in Eqs. 76, 77.

The two functions must match at x = y, otherwise G(x,y) would have a discontinuity there,
with a δ(x−y) in its derivative. This gives the condition

αy = β(L−y) (69)

which we satisfy by setting

α = c(L−y) (70)
β = cy (71)

with c a constant still to be determined. If we take the derivative of G we now find

dG(x,y)

dx

= c(L−y) for x < y (72)

dG(x,y)

dx
= −cy for x > y (73)

with a discontinuity

∆dG/dx =
dG(x,y)

dx

∣∣∣
y+
−
dG(x,y)

dx

∣∣∣
y−

= −cy − c(L−y) = −cL (74)

The discontinuity in dG/dx at x = y implies

d2G(x,y)

dx2
= −cLδ(x−y) (75)

10

Equation 63 demands then cL = 1 fixing the value of c, namely fixing c = 1/L. We thus
obtain for the Green’s function

G(x,y) =
x(L−y)

L
for x < y (76)

G(x,y) =
(L−x)y

L
for x > y (77)

Note that G(x,y) is symmetric under the exchange x ↔ y.

Solution with the Fourier series.

We considered above the Fourier series for functions defined over −L/2 ≤ x ≤ L/2 with
periodic boundary conditions. This can be easily generalized to functions f(x) defined over
0 ≤ x ≤ L and Dirichlet boundary conditions f(0) = f(L) = 0. In this case the Fourier
series takes the form

f(x) =

∞∑
n=1

Fn

√
2

L
sin

πnx

L
(78)

The normalization in Eq. 78 insures that∫ L
0

f∗(x)f(x) dx =

∫ L
0

(

∞∑
m=1

F∗m

√
2
L
sin

πmx

L

)

( ∞∑
n=1

Fn
√
2
L
sin
πnx
L

)
dx =

∞∑
m=1
∞∑
n=1

F∗mFn
2

L
∫ L
0

sin
πmx

L
sin
πnx

L
dx =

∞∑
m=1
∞∑
n=1
F∗mFn
2
L
L

2
δm.n =

∞∑
n=1

F∗nFn (79)

where we used the othonormality relation

2
L
∫ L
0
sin
πmx
L
sin
πnx

L
dx = δm,n (80)

The Fourier coefficients Fn can be obtained from f(x) by

Fn =

√
2
L
∫ L
0

f(x)

sin
πnx

L
dx (81)

Inserting the expression for f(x) given by Eq. 78 and using again the orhonormality relation
Eq. 80 one can easily check that the r.h.s. of Eq. 81 is indeed equal to Fn.

Let now f(x) be the solution of Eq. 61 and let the Fourier series expansion of b(x) be

b(x) =
∞∑
n=1

Bn

√
2
L
sin
πnx

L
(82)

Substituting Eqs. 78 and 82 into Eq. 61 we find

−
d2

dx2
( ∞∑
n=1
Fn
√
2
L
sin
πnx
L

)
=
∞∑
n=1

(πn
L

)2
Fn

√
2
L
sin
πnx

L
=

∞∑
n=1

Bn
√
2
L
sin
πnx

L
(83)

11

from which we deduce

Fn =

( L
πn

)2
Bn (84)

Inserting this result into Eq. 78 we get

f(x) =
∞∑
n=1

( L
πn

)2 √ 2
L

sin
πnx

L
Bn (85)

and using (see Eq. 81)

Bn =

√
2
L
∫ L
0

b(y) sin

πny

L
dy (86)

we further obtain

f(x) =
∫ L
0
∞∑
n=1

2L

(πn)2
sin

πnx
L
sin
πny

L
b(y) dy (87)

where we exchanged the integral with the sum and rearranged a few terms. If we define

G(x,y) =
∞∑
n=1

2L
(πn)2
sin
πnx
L
sin
πny

L
(88)

then Eq. 87 takes the form

f(x) =
∫ L
0

G(x,y)b(y) dy

But this is precisely the equation, Eq. 64, which defines the Green’s function. So we conclude
that Eq. 88 gives an alternative form of the Green’s function, expressed now as a Fourier
series.

The challenge.

We have two expressions for the Green’s function: Eqs. 76, 77

G(x,y) =
x(L−y)

L
for x < y

G(x,y) =
(L−x)y

L
for x > y

and Eq. 88

G(x,y) =
∞∑
n=1
2L
(πn)2
sin
πnx
L
sin
πny
L

The challenge is to prove that they are the same function. Or, more precisely, we want to
sum the series in Eq. 88 and prove that the result of the sum coincides with Eqs. 76, 77. We
can sum the series by expressing it in terms of a contour integral in the complex plane.

12

Consider the integral

I =

∫
c1

F(z)

sin z
dz (89)

where c1 is a contour consisting of a line parallel to the real axis, but displaced by a small
amount � above it, from +∞ down to a value between 0 and π, e.g. down to π/2; the contour
then makes a small counterclockwise half loop around π/2 and proceeds back to +∞ along a
line parallel to the real axis, but displaced by a small amount � below it. The function F(z)
is supposed to be analytic in a domain encompassing the contour and the region enclosed
by it. The integrand has simple poles for z = πn with n = 1, 2, . . . and is otherwise analytic
inside the c1. Thus the integral is equal to 2πı times the sum of the residues

I = 2πı
∞∑
n=1

(−1)nF(πn) (90)

We can get rid of the oscillating factor (−1)n by multiplying F(z) by either exp(ız) or
exp(−ız). It is good to maintain the freedom of using either sign in the exponential. Thus
we find

∞∑
n=1

F(πn) =
1

2πı
∫
c1

F(z)e±ız

sin z
dz (91)

Consider another contour c2 consisting of a line parallel to the real axis, but displaced
by a small amount � below it, from −∞ up to −π/2; the contour then makes a small
counterclockwise half loop around −π/2 and proceeds back to −∞ along a line parallel to
the real axis, but displaced by a small amount � above it. The function F(z) is supposed
to be analytic in a domain encompassing the contour and the region enclosed by it. The
integrand has simple poles for z = −πn with n = 1, 2, . . . and is otherwise analytic inside
the c2. Proceeding in a manner analogous to the above we find

−∞∑
n=−1

F(πn) =
1
2πı

∫
c2

F(z)e±ız

sin z
dz (92)

If the function F(z) is even under z →−z, i.e. if F(−z) = F(z), we can add the two integrals
to obtain

∞∑
n=1
F(πn) =
1

4πı

∫
c1+c2

F(z)e±ız

sin z
dz (93)

or, with

F(z) =
2L

z2

(
sin

zx

L
sin

zy

L

)
(94)

13

γ

Figure 2: Contour integral used in the calculation of the Green’s function.

∞∑
n=1
2L
(πn)2
sin
πnx
L
sin
πny
L
=
1
4πı
∫
c1+c2
2L

z2 sin z

(
sin
zx
L
sin
zy

L
e±ız

)
dz =

−
L

8πı

∫
c1+c2
1
z2 sin z

(
eızx/L −e−ızx/L

)(
eızxy/L −e−ızy/L

)
e±ız dz =

−
L
8πı
∫
c1+c2
1
z2 sin z

(
eız(x+y)/L + e−ız(x+y)/L −eız(x−y)/L −e−ız(x−y)/L

)
e±ız dz (95)

The strategy is now to close the contour c1 + c2 with two big half-circles of large radius R
in the upper and lower half planes, making the whole a big clockwise loop γ. If as R → ∞
the integrand in Eq. 95 goes to zero faster than 1/R then the integral over c1 + c2 can be
replaced by an integral over the closed contour γ. It is actually convenient to think of γ as
the same loop in the counterclockise direction, which simply involves a change of sign in the

14

integral which now takes the form

L
8πı

∮
γ

1
z2 sin z
(
eız(x+y)/L + e−ız(x+y)/L −eız(x−y)/L −e−ız(x−y)/L

)
e±ız dz (96)

The only singularity of the integrand inside γ is at z = 0 and we will be able to calculate
the integral with the theorem of residues. We must however make sure that the integrand
vanishes fast enough for R →∞ and this is where the ability to use either exp(ız) or exp(−ız)
to remove the sign (−1)n comes into play. Let us begin by considering the denominator.
Since sin z contains both a term with exp(ız) and a term with exp(−ız) it will grow as exp(R)
whether z = ıR (upper half plane) or z = −ıR (lower half plane). So, including also the
z2 factor, the denominator will grow as R2 exp(R) and will contribute a factor decreasing
as exp(−R)/R2 to the integrand. But we have to be careful that the four terms in the
numerator do not grow in a manner that will overcome the exp(−R) suppression. This is
where it will become crucial to distinguish between the two cases x ≤ y and x ≥ y, which
play an important role in the Green’s function formulae of Eqs. 76, 77. So let us take for
definiteness x ≥ y (the other case can be dealt with simply interchanging x and y.) We must
consider the behavior of the four terms in the numerator separately.

1 : eız(x+y)/L (97)

(x + y)/L will never exceed 2. So if we use exp(−ız) generating a term

f1(z) = e
ız[(x+y)/L−1] (98)

we will be sure that the term never grows faster than exp(R) (in this case for z = −ıR) and
we will be o.k.

2 : e−ız(x+y)/L (99)

(x + y)/L will never exceed 2. So if we use exp(ız) generating a term

f2(z) = e
−ız[(x+y)/L−1] (100)

we will be sure that the term never grows faster than exp(R) (in this case for z = ıR) and
we will be o.k.

3 : eız(x−y)/L (101)

This is where x ≥ y plays a role. (x − y)/L will never be larger than 1 and will never be
negative, since x ≥ y. Thus if we use exp(−ız) generating a term

f3(z) = e
ız[(x−y)/L−1] (102)

we will be sure that the term never grows faster than exp(R) and we will be o.k. Similarly
for

4 : e−ız(x−y)/L (103)

15

we will use exp(ız) generating a term

f4(z) = e
−ız[(x−y)/L−1] (104)

We can now apply the theorem of residues. The numerator in Eq. 96 has a zero of degree
2 at the origin, as is especially clear from its sin(zx/L) sin(zy/L) form in Eq. 95. The
denominator has a zero of degree 3 at the origin. So the whole integrand has a simple pole
there. To make the residue explicit we expand

f1(z) + f2(z) −f3(z) −f4(z) =
eız[(x+y)/L−1] + e−ız[(x+y)/L−1] −eız[(x−y)/L−1] −e−ız[(x−y)/L−1] =

−z2[(x + y)/L− 1]2 + z2[(x−y)/L− 1]2 + · · · =
−z2[2x/L− 2][2y/L] = 4z2(L−x)y/L2 + . . . (105)

where the dots represent higher order terms which can be neglected since they do not con-
tribute to the singular part of the integrand. With this the integral in Eq. 96 becomes

(L−x)y
L

1
2πı
∮
γ
1

sin z
dz =

(L−x)y
L

(106)

in agreement with Eq. 77. The expression for x ≤ y is obtained by simply interchanging x
and y.

16