a companion to fourier analysis for physics students
TRANSCRIPT
A Companion to Fourier Analysis
for Physics Students
(Part One: Fourier Series)
Gol Mohammad Nafisi∗
University of Tehran
Spring 2019
Contents
1 Prelude 1
2 Preliminaries 2
3 Fourier Trigonometric Series 3
4 Operations on Fourier Series 7
5 Complex Fourier Series 11
6 Fourier Series in Action 14
6.1 Electrostatic Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
6.2 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
7 Coda 19
i
1 Prelude
بصیر چشم فایده بود چه نبین گر است نظر برای مطبوع ر پی سعدیا
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
As a caring TA1, I’ve written these short companion notes with the intention of preparing you guys for do-
ing essential calculations you might encounter in any of your undergraduate Physics courses regarding the topic
of Fourier analysis (fingers crossed!). To do so, we are going to review some essential tools of the trade and then
along the road we are going to do a fair amount of examples, since one does not simply learn mathematics just by
memorizing theorems and definitions. I do not claim on originality of the material whatsoever. As David Tong
put it, “My primary contribution has been to borrow, steal and assimilate the best discussions and explanations I
could find from the vast literature on the subject”.
I have divided the subject into two parts. In part one, we are going to deal with Fourier series and part two will
contain the Fourier transform. I should remind you that I have prepared these notes in order to be used in our
weekly TA recitation sessions and they are NOT going to be a replacement of any sort for the class lectures or
the course reference textbooks. I am positive that together we will have a joyful ride and as Sidney Coleman said
once, “Not only God knows, I know, and by the end of the semester, you will know”. Last but not least, any
questions or comments are very welcome. Now let’s begin, shall we?
1Yeah, that’s the spirit.
1
2 Preliminaries
“You can’t suddenly know something just by assembling a committee of words! That’s it! I’ll assemble your committee.”
- Professor Farnsworth
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Historically, one can trace the origin of Fourier analysis back to a breakthrough in the investigations regarding the
problem of vibrating string and heat equation made by Joseph Fourier. His bold insight was that we can model
all functions by trigonometric series, which then became known as the Fourier series. In other words, all functions
can be expanded as the superposition of basic (Sine and Cosine) waves with definite frequencies. In general, we
have two types of Fourier expansions: Fourier series and Fourier transform.
In Fourier series, we deal with periodic functions which can be written as a discrete sum of trigonometric (or
exponential) functions with definite frequencies. In Fourier transform, they do not need to be periodic and they
can be represented by a continuous integral of trigonometric (or exponential) functions with a continuum of
possible frequencies. Now let’s review some stuff first.
• Consider a function f : X → Y , for which ∃T > 0 , s.t.
(a) ∀x ∈ X : x± T ∈ X
(b) ∀x ∈ X : f(x± T ) = f(x)
Then f(x) is periodic and T is its (fundamental) period. For example, the fundamental period of the
functions sinn(ax) and cosn(ax) is T =π
|a|if n is even and T =
2π
|a|if n is odd.
• The function f is odd if f(−x) = −f(x) and it is even if f(−x) = f(x) . Also f can be neither e.g. f(x) = ex .
• For functions that are not intrinsically periodic, we can define a period. For example, one can define a period
T = 2π for f(x) = x where x ∈ [−π, π] .
• These identities might come in handy:
(a) ∀n ∈ N :
• cos(nπ) = (−1)n
• sin(nπ) = 0
• sin((2n− 1)
π
2
)= (−1)n−1
(b) sin(α± β) = sinα cosβ ± cosα sinβ
(c) cos(α± β) = cosα cosβ ∓ sinα sinβ
2
3 Fourier Trigonometric Series
“I wish there was a way to know you’re in the good old days before you’ve actually left them.”
- Andy Bernard
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
In this section we will first introduce the Fourier theorem and work it through examples.
Theorem 1 (Fourier Series). Consider a function f(x) with the period of T = 2L (e.g. x ∈ [−L,L]) . Then we
can represent f by:
f(x) = a0 +
∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)](1)
where
a0 =1
2L
∫ L
−Lf(x) dx
an =1
L
∫ L
−Lf(x) cos
(nπxL
)dx
bn =1
L
∫ L
−Lf(x) sin
(nπxL
)dx
(2)
Remark 1. For odd functions we have a0 = an = 0 and for even functions we have bn = 0 .
Example 1. Let
f (x) =
−k −π < x < 0
k 0 < x < π
(3)
Write down the trigonometric Fourier series of f .
Solution 1. First
a0 =1
2π
∫ π
−πf (x) dx =
1
2π
∫ 0
−πf (x) dx+
1
2π
∫ π
0f (x) dx
=1
2π
∫ 0
−π(−k) dx+
∫ π
0(k) dx =
1
2π{(−k) (0 + π) + (k) (π − 0)} = 0
(4)
Nextan =
1
π
∫ π
−πf (x) cos (nx) dx =
1
π
{∫ 0
−π(−k) cos (nx) dx+
∫ π
0(k) cos (nx) dx
}=
1
π
{(−k) 1
n[sin (nx)]−π
0 + (k)1
n[sin (nx)]π0
}= 0
(5)
3
And
bn =1
π
{∫ 0
−π(−k) sin (nx) dx+
∫ π
0(k) sin (nx) dx
}=
1
π
{(−k) 1
n[−cos (nx) ]0−π + (k)
1
n[−cos (nx) ]π0
}
=2k
nπ[1− cos (nπ) ] = 2k
nπ[1− (−1)n] =
0 even n
4k
nπodd n
(6)
Finally, we can substitute the obtained results in the general formula for the Fourier series and obtain
f(x) =
∞∑n=1
4k
(2n− 1)πsin [(2n− 1)x] (7)
Remark 2. Sometimes it’s useful for an even function, to change the integration interval from [−L,L] to [0, L]
and then multiply the integral by the factor of 2.
Example 2 (Arfken 19.1.1). Using the standard definition of derivatives, we obtain:
∂∆p
∂an=
∫ 2π
02{f (x)− a0
2−
p∑n=1
[ancos (nx) + bnsin (nx) ] [−cos (nx) ]} dx
= −2{∫ 2π
0f (x) cos (nx) dx−
∫ 2π
0
a02
cos (nx) dx−p∑
n=1
∫ 2π
0ancos (nx) cos (nx) dx
−p∑
n=1
∫ 2π
0bnsin (nx) cos (nx) dx
}(8)
The orthogonality of Sine and Cosine functions simplifies the third and fourth integrals as:
∫ 2π
0ancos (nx) cos (nx) dx = πan∫ 2π
0bnsin (nx) cos (nx) dx = 0
(9)
Hence∂∆p
∂an= 0 = −2
∫ 2π
0f (x) cos (nx) dx+ 2πan =⇒ an =
1
π
∫ 2π
0f (x) cos (nx) dx (10)
Same procedure goes for the bn therefore the statement is true.
4
Example 3. We’ll write down the Fourier expansion of Dirac delta function δ(x) . Delta function is not intrinsically
periodic, but if we let the period be T = 2π and use the property∫ +∞
−∞f (x) δ (x− x0) dx = f(x0) we get:
a0 =1
2π
∫ π
−πδ (x) dx =
1
2π
an =1
π
∫ π
−πδ (x) cos (nx) dx =
1
π[cos (0)] = 1
π⇒ δ (x) =
1
2π+
1
π
∞∑n=1
cos(nx)
bn =1
π
∫ π
−πδ (x) sin(nx)dx =
1
π[sin (0)] = 0
(11)
Remark 3. Sometimes for a specific value of n , the denominator of an or bn vanishes. In this case we need to
calculate the coefficients for this specific value directly from the formulas. Following example will shed light on
this remark.
Example 4. We want to write down the Fourier expansion of the following function:
f(x) =
sin(x) 0 < x < π
0 −π < x < 0
(12)
Our solving algorithm includes the following steps:
(1) Finding the period and integration interval:
T = 2π → L = π (13)
(2) Finding the zeroth coefficient:
a0 =1
2π
{∫ 0
−π(0) dx+
∫ π
0sin (x) dx
}=
1
π(14)
(3) Finding the general coefficient an :
an =1
π{∫ 0
−π
{(0) cos (nx) dx+
∫ π
0sin (x) cos (nx) dx
}=
1
π
{∫ π
0
1
2[sin [(1− n)x] + sin [(1 + n)x]] dx
}=
1
2π
{[−cos (1− n)x
1− n
]π0
+
[−cos (1 + n)x
1 + n
]π0
}=
1
2π
{cos (nπ)1− n
+cos (nπ)1 + n
+1
1− n+
1
1 + n
}→ an =
1 + cos(nπ)π(1− n2)
(15)
5
Now you see that for n = 1 , coefficient an is ill-defined therefore we must compute a1 separately from
the general formula:
a1 =1
π
∫ π
0sin (x) cos (1× x) dx =
1
2π
∫ π
0sin (2x) dx = 0 (16)
(4) Finding bn :
bn =1
π
{∫ 0
−π(0) sin (x) dx+
∫ π
0sin (x) sin (nx) dx
}=
1
π
∫ π
0
1
2{cos [(1− n)x]− cos [(1 + n)x]} dx
=1
2π
[sin (1− n)x
1− n
]π0
+1
2π
[−sin (1 + n)x
1 + n
]π0
= 0
(17)
Again, we need to calculate b1 separately:
b1 =1
π
∫ π
0sin (x) sin (1× x) dx =
1
π
∫ π
0
1− cos(2x)2
dx =1
2(18)
(5) Final step includes putting together all the results into a neat expression:
f (x) =1
π+
1
2sin (x) +
1
π
∞∑n=2
[1 + cos (nπ)
1− n2
]cos(nx) (19)
6
4 Operations on Fourier Series
“Go ahead, make my day.”
- Harry Callahan
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
In this section we will re-learn some tricks which will come handy in some situations. First let me review some
concepts2.
Definition 1. A function is called piecewise continuous on an interval if the interval can be broken into a
finite number of sub-intervals on which the function is continuous on each open sub-interval (i.e. the sub-interval
without its endpoints) and has a finite limit at the endpoints of each sub-interval.
Definition 2. We say that the function f(x) is piecewise smooth if the function can be broken into distinct
pieces and on each piece both the function and its derivative, f ′(x) , are continuous. A piecewise smooth function
may not be continuous everywhere however the only discontinuities that are allowed are a finite number of jump
discontinuities.
Theorem 2 (Term-by-term differentiation of Fourier series). If f(x) is a piecewise smooth function and
if f(x) is also continuous on [−L,L] , then the Fourier series of f(x) can be term-by-term differentiated if
f (−L) = f (L) .
Theorem 3 (Term-by-term integration of Fourier series). Fourier series of a piecewise smooth function
f(x) can always be term-by-term integrated to give a convergent series that always converges to the integral of
f(x) , ∀x ∈ [−L,L] .
Example 5. Let the Fourier series of f(x) = | sin(x)| where −π < x < π be given by
f(x) =2
π+
4
π
∞∑n=1
cos(2nx)1− 4n2
(20)
Find the Fourier expansion of f(x) = cos(x) .
Solution 5. By taking the derivative of both sides of eq.20 for f(x) = sin(x) on the interval 0 < x < π we get
cos(x) = − 8
π
∞∑n=1
n sin(2nx)1− 4n2
(21)
2I’ve borrowed the first two definitions from Paul Dawkins’ cool notes available at http://tutorial.math.lamar.edu
7
Example 6. Let the Fourier series of function
f(x) =
−1 −π < x < 0
+1 0 < x < π
(22)
be given by
f(x) =4
π
{sin(x) + sin(3x)
3+
sin(5x)5
+ · · ·}
(23)
Find the Fourier series of function
g(x) = |x| where − π < x < π (24)
Solution 6. If you look carefully, you’ll see that g(x) =∫f(x) dx. Hence
g(x) =4
π
∫ x
0
[sin(t) + sin(3t)
3+ · · ·
]dt =
4
π
∞∑n=0
∫ x
0
sin((2n+ 1) t)
2n+ 1dt =
4
π
∞∑n=0
1− cos((2n+ 1)x)
(2n+ 1)2(25)
Now fun stuff begins by introducing following theorems and utilizing them to find the numerical values of some
infinite series. Look out! Riemann Zeta Function ahead.
Theorem 4 (Dirichlet’s Theorem). Suppose f(x) is a piecewise smooth function on the interval −L ≤ x ≤ L ,
then ∀x ∈ [−L,L] :
a0 +∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]=
1
2
[f(x+) + f(x−)
](26)
where f(x+) and f(x−) are the right/left limits of f respectively.
Theorem 5 (Parseval’s Theorem). Let a0 , an and bn be the Fourier coefficients of the periodic square-integrable
function f(x) over −L < x < L . Then the following identity holds
2a20 +∞∑n=1
(a2n + b2n
)=
1
L
∫ L
−L[f(x)]2 dx (27)
8
Example 7. Let the Fourier expansion of function
f(x) = 4− x2 where − 2 ≤ x ≤ 2 (28)
be given by
f(x) =8
3+
16
π2
{cos(πx
2
)− 1
4cos (πx) + 1
9cos(3πx
2
)− · · ·
}(29)
Calculate ζ(2) , where ζ(n) is the Riemann Zeta Function defined by
ζ(n) =
∞∑k=1
1
kn(30)
Solution 7. Since f is continuous at x = 2 then due to Dirichlet’s theorem, the value of f is equal to the value
of its Fourier series at this point. First:
f(x) = 4− x2 ⇒ f(2) = 0 (31)
Then
f(2) =8
3+
16
π2
{cos(π × 2
2
)− 1
4cos (π × 2) +
1
9cos(3π × 2
2
)− · · ·
}=
8
3+
16
π2
{−1− 1
22− 1
32− · · ·
}(32)
Hence
0 =8
3− 16
π2
{1 +
1
22+
1
32+ · · ·
}⇒ ζ(2) =
∞∑k=1
1
k2=π2
6(33)
Example 8. Let the Fourier Cosine series of f(x) = sin(x) where 0 ≤ x ≤ π be
f(x) =2
π− 2
π
∞∑n=1
[1 + cos(nπ)n2 − 1
]cos(nπx) (34)
Evaluate the numerical value of the following infinite series:
S =∞∑k=1
1
(2k − 1)2 (2k + 1)2(35)
Solution 8. Fourier Cosine series of f(x) over 0 ≤ x ≤ L is given by
f(x) = a0 +∞∑n=1
an cos(nπxL
)s.t. a0 =
1
L
∫ L
0f(x) dx ; an =
2
L
∫ L
0f(x) cos
(nπxL
)dx (36)
9
Now from the given expansion we see that
a0 =2
π; an = − 2
π
[1 + cos(nπ)n2 − 1
]=
0 odd n
− 2
π
(2
n2 − 1
)even n
(37)
Using Parseval’s theorem we get (L = π) :
2
(2
π
)2
+
∞∑n=1
[− 2
π
(2
n2 − 1
)]2=
1
π
∫ π
−πsin2(x) dx (38)
Letting n = 2k we get8
π2+
16
π2
∞∑k=1
(1
4k2 − 1
)2
=2
π
∫ π
0
[1− cos(2x)
2
]dx = 1 (39)
Hence∞∑k=1
1
(2k − 1)2 (2k + 1)2=π2
16
(1− 8
π2
)(40)
10
5 Complex Fourier Series
“Of course it is happening inside your head, Harry, but why on earth should that mean that it is not real?”
- Professor Dumbledore
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Any arbitrary periodic function that can be written in terms of trigonometric Fourier series, can also be written
as a complex exponential form:
f (x) =+∞∑
n=−∞Cn e
inπxL where Cn =
1
2L
∫ L
−Lf (x) e−
inπxL dx (41)
Remark 4. Relationship between complex coefficients and real ones is given by
an = Cn + C−n ; bn = i (Cn − C−n) (42)
Example 9. Find the complex Fourier series of the periodic function f (x) = x2 , ∀x ∈ [0, 2π] .
Solution 9. First we have
Cn =1
2π
∫ 2π
0x2 e−inx dx (43)
As you might recall from Calculus, integrals of this kind can be calculated via integration by parts. For the sake
of fanciness, the general formula is3:
∫xn eax dx = eax
(n∑
k=0
(−1)k k!
(n
k
)xn−k
ak+1
)(44)
Hence
Cn =1
2π
[e−inx
(− 2i
n3+
2x
n2+ix2
n
)]2π0
=2 + 2iπn
n2(45)
You see that for n = 0 we have an ill-defined quantity, so we calculate C0 from the general formula:
C0 =1
2π
∫ 2π
0x2 dx =
4π2
3(46)
Therefore
f(x) =4π2
3+
∞∑n=−∞n̸=0
2 + 2iπn
n2einx (47)
3Table of Integrals, Series, and Products, 7th Edition, I.S. Gradshteyn and I.M. Ryzhik, Academic Press
11
Example 10. Write down the complex Fourier series of the following function:
f (x) =a sinx
1− 2a cosx+ a2s.t. |a| < 1 (48)
Solution 10. “Don’t be intimidated, Squidward!” Try to use these formulas:
sinx =eix − e−ix
2i; cosx =
eix + e−ix
2(49)
Hence
f (x) =a2i
(eix − e−ix
)1− 2a eix+e−ix
2 + a2=
1
2i
a(eix − e−ix)
(1− aeix)(1− ae−ix)(50)
Now you recall the partial-fraction decomposition? Apply it here to get:
f (x) =1
2i
a(eix − e−ix)
(1− aeix)(1− ae−ix)=
1
2i
{A
1− aeix+
B
1− ae−ix
}→
A = 1
B = −1
⇒ f (x) =1
2i
[1
1− aeix− 1
1− ae−ix
] (51)
Recall the Laurent series of the function f(z) =1
1− z. There was a condition to represent it as a convergent
series. What was it? Yes! It was |z| < 1 . Now we know that∣∣a e±ix
∣∣ = |a| and from the statement of the problem
we know that |a| < 1, so the condition is satisfied and we can expand f(x) in the form of power series where:
1
1− aeix=
∞∑n=0
an einx ;1
1− ae−ix=
∞∑n=0
an e−inx (52)
Hence
f (x) =1
2i
∞∑n=0
an(einx − e−inx
)=
∞∑n=0
an sin(nx) =∞∑n=1
an sin(nx) (53)
Example 11 (Arfken 19.1.4). Assuming that∫ π
−π[f(x)]2 dx is finite, show that
limn→∞
an = 0 , limn→∞
bn = 0 (54)
Solution 11. By Parseval’s theorem we get
1
π
∫ π
−π[f(x)]2 dx =
1
2a20 +
∞∑n=1
(a2n + b2n
)(55)
12
Since from the statement LHS is finite, then the RHS must converge and we know that the series∞∑n=1
An is
convergent if
limn→∞
An = 0 (56)
Example 12 (Arfken 19.1.15). Given
ψ2s(x) =∞∑n=1
sin(nx)n2s
, ψ2s+1(x) =∞∑n=1
cos(nx)n2s+1
(57)
develop the following recurrence relations:
(a) ψ2s(x) =
∫ x
0ψ2s−1(x) dx (b) ψ2s+1(x) = ζ(2s+ 1)−
∫ x
0ψ2s(x) dx
Solution 12.
(a) Since d
dxψ2s =
∞∑n=1
cos (nx)n2s−1
it seems proper that we let 2s+ 1 7→ 2s− 1 in ψ2s+1 hence:
ψ2s−1 =∞∑n=1
cos (nx)n2s−1
→∫ x
0ψ2s−1dx =
∫ x
0
∞∑n=1
cos (nx)n2s−1
dx =∞∑n=1
sin (nx)
n2s= ψ2s (58)
(b) We have ∫ x
0ψ2s dx =
∫ x
0
∞∑n=1
sin (nx)
n2sdx = −
∞∑n=1
cos (nx)n2s+1
+
∞∑n=1
1
n2s+1(59)
Recall the Riemann Zeta Function ζ(2s+ 1) =∞∑n=1
1
n2s+1, hence
ψ2s+1 = ζ (2s+ 1)−∫ x
0ψ2s dx (60)
13
6 Fourier Series in Action
“Admiration is the emotion furthest from understanding.”
- Sosuke Aizen
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
One of the frequent places in physics where Fourier series appears is when we want to determine the unknown
coefficients of the general solution of a PDE. As you might recall from the Linear Algebra lectures, we had the
following theorem
Theorem 6. Suppose that {ϕn}∞n=0 is an orthogonal set of functions on [a, b] with respect to the weight function
w(x) . If f(x) is a function on [a, b] and
f(x) =
∞∑n=0
an ϕn(x) (61)
then the coefficients an are given by
an =〈f, ϕn〉〈ϕn, ϕn〉
=
∫ ba f(x)ϕn(x)w(x) dx∫ ba [ϕn(x)]
2 w(x) dx(62)
Now letting some rigor in, all the stuff we talked about Fourier series lies on the fact that the functions must be
elements of L2 space:
Definition 3. L2 ([−a, a]) is the set of all complex-valued functions on [−a, a] satisfying
∫ a
−a|f(x)|2 dx <∞ (63)
with the inner product
〈f, g〉 = 1
a
∫ a
−af(x) g(x) dx (64)
These are the so-called square-integrable functions on [−a, a] .
I hope you can see the big picture now. Since the functions {sin(nx), cos(nx)} and {eikx : k ∈ Z} form the
orthogonal sets of functions, the Fourier series becomes just an orthogonal expansion for some periodic square-
integrable functions and all the Fourier coefficients are just like the ones we saw in previous theorem where the
weight function is w(x) = 1 .
In most cases where we deal with Laplace, Heat or Wave equations, applying the initial or boundary conditions
will enable us to use the mentioned remark in order to find the unknown coefficients of the general solution. Eager
to see how it works?! Then buckle up.
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6.1 Electrostatic Potential
Let’s start with an example about how can we use Fourier series in the realm of electrostatics. The main goal here
is to find a suitable potential (function) which satisfies some conditions. You can think of a potential here as a
proper solution of the Laplace (or Poisson) equation. It might be interesting to know that the twice continuously
differentiable functions which satisfy Laplace equation are called harmonic functions. Recall that the Laplace
equation is in fact a PDE which is independent of time (meaning that we have no initial conditions), whose solu-
tions can be uniquely obtained by applying specific boundary conditions (values of the solution on some surfaces
or regions of space). We’ll use the separation of variables method to solve the PDEs here.
Example 13. Solve the following boundary-value problem inside a rectangular of length a and width b:
∇2U (x, y) = 0 ; 0 < x < a , 0 < y < b
U (0, y) = 0 ; 0 ≤ y ≤ b
U (x, 0) = 0 ; U (x, b) = f (x) ; 0 ≤ x < a
(65)
Solution 13. Let’s draw a picture first4.
Let’s assume that the general solution has the form of U (x, y) = X (x)Y (y) . Substituting this into the PDE
yields:X
′′(x)
X(x)+Y
′′(y)
Y (y)= 0 (66)
Assuming that two fractions have the same value (separation constant), we get:
X′′(x)
X(x)= −λ , Y
′′(y)
Y (y)= λ (67)
Applying the boundary conditions yields: U (0, y) = X (0)Y (y) = 0 → X (0) = 0
4Since they say “A picture is worth a thousand words”.
15
and U (a, y) = X (a)Y (y) = 0 → X (a) = 0 . Therefore for x, we have the following ordinary differential equation:
X
′′(x) + λX (x) = 0
X (0) = X (a) = 0
(68)
Letting λ = ω2 where ω ∈ R , gives us the following familiar solution:
X (x) = A cos (ωx) +B sin(ωx) (69)
The boundary conditions determine the eigenvalues and eigenfunctions as: Xn (x) = Bn sin(nπx
a
). Now using
the obtained eigenvalues, we have the following solution for Y (y) 5:
Yn (y) = Cn cosh(nπy
a
)+Dn sinh
(nπya
)B.C.==⇒ Yn (y) = Dn sinh
(nπya
)(70)
Hence the general solution will be:
Un (x, y) =
∞∑n=1
En sinh(nπy
a
)sin(nπx
a
)(71)
Now let’s apply the final boundary condition:
U (x, b) = f (x) → f (x) =
∞∑n=1
En sinh(nπb
a
)sin(nπx
a
)(72)
Now do you see any familiar stuff here? You’re right! f has expanded as the Fourier Sine series with the coefficients
bn ≡ En sinh(nπb
a
). You do remember that for the Fourier coefficients we had bn =
2
L
∫ L
0f (x) sin
(nπxL
)dx ,
hence here for L = a we get
En sinh(nπb
a
)=
2
a
∫ a
0f (x) sin
(nπxa
)dx → En =
2
a sinh(nπba
) ∫ a
0f (x) sin
(nπxa
)dx (73)
Voila! You can see that if the explicit form of f(x) is given, then you can calculate En , ergo finding the explicit
form of Un (x, y), i.e. the solution of Laplace equation for this problem.
Whoa! What a lengthy answer for such a short problem! But don’t worry, you’ll get used to it by more practicing,
or maybe you won’t. Anyway, let’s get physical by the next example.
5You can use the exponential form of the solution as well.
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Figure 1: Cross section of split tube
Example 14 (Arfken 19.2.11). A metal cylindrical tube of radius a is split lengthwise into two non-touching
halves (Fig.1). The top half is maintained at a potential +V , the bottom half at a potential −V . Separate the
variables in Laplace equation and solve for the electrostatic potential for r ≤ a .
Solution 14. The recipe for this problem is same as the previous one, except that the geometry has changed and
we need to work in cylindrical coordinates. The general form of the Laplace equation in cylindrical coordinates is
∇2ϕ (r, φ, z) =1
r
∂
∂r
(r∂ϕ (r, φ, z)
∂r
)+
1
r2∂2ϕ (r, φ, z)
∂φ2+∂2ϕ (r, φ, z)
∂z2= 0 (74)
By separation method, we have ϕ (r, φ, z) = R (r)F (φ)Z(z) . Substituting this into Laplace equation yields:
1
R
d2R
dr2+
1
rR
dR
dr+
1
r2F
d2F
dφ2 +1
Z
d2Z
dz2= 0 (75)
Let 1
Z
d2Z
dz2= k2 where k ∈ R . This will give us the equation for Z(z) as d
2Z
dz2− k2Z (z) = 0 whose solution is
Z(z) = e±kz . Now let 1
F
d2F
dφ2 = −m2 → d2F
dφ2 +m2F (φ) = 0 ⇒ F (φ) = Am cos (mφ) + Bm sin (mφ) . The only
remaining ODE isd2R
dr2+
1
r
dR
dr+
(k2 − m2
r2
)R = 0 (76)
Not an easy one it seems. But if we let x = kr then d2R
dx2+
1
x
dR
dx+
(1− m2
x2
)R = 0 , which is the Bessel equation
whose solutions are called Bessel functions of order m6. For the sake of space and time, let’s assume that the
potential is symmetric in z, i.e. ϕ (r, φ, z) ≡ ϕ (r, φ) = R (r)F (φ) . Therefore the radial equation takes the form of
r2R′′(r)+rR′ (r)−m2R (r) = 0 , which has the power series solution of the form R (r) =∞∑
m=1
{Cm r
m +Dm r−m}
.
Hence
ϕ (r, φ) =∞∑
m=1
{Cm r
m +Dm r−m}{Am cos (mφ) +Bmsin (mφ)} (77)
Now since our region of interest includes the origin (r = 0) , then we require our solution be regular at this point.6For further information on these creatures, see e.g. Alan Jeffrey & Hui-Hui Dai’s Handbook of Mathematical Formulas and Integrals
17
Therefore we would omit the r−m terms. Also, you can observe that the potential is an odd function with respect
to angle φ 7, hence Am = 0 . Applying the boundary conditions yield:
ϕ (a, φ) = ±V =∞∑
m=1
amBm sin (mφ) → amBm =1
π
∫ 2π
0(±V ) sin (mφ) dφ
=V
π
{∫ π
0sin (mφ) dφ−
∫ 2π
πsin (mφ) dφ
}=
4V
mπs.t. m is odd
(78)
Therefore
ϕ (r, φ) =4V
π
∞∑n=0
(ra
)2n+1 sin((2n+ 1)φ)
2n+ 1(79)
6.2 Heat Transfer
We can apply the same procedure of finding unknown coefficients in the case of heat equation as well.
Example 15. Suppose that we have a metal rod of length 10m whose endpoints are kept at zero temperature
and at time t = 0 , its temperature was u(x, 0) = x (10− x) . If the temperature of rod is given by
u(x, t) =∞∑n=1
An sin(√
λnx)e−λn c2 t s.t. λn =
(nπ10
)2, c2 = 1 (80)
by determining the unknown coefficients, calculate the temperature at midpoint of the rod.
Solution 15. By initial condition we have
x (10− x) =
∞∑n=1
An sin(nπx
10
)(81)
You see that An is the Fourier Sine coefficient of x (10− x) . Hence
An =2
10
∫ 10
0x (10− x) sin
(nπx10
)dx =
400
n3π3[1− (−1)n] (82)
Therefore
u(x, t) =∞∑n=0
800
(2n+ 1)3π3sin((2n+ 1)πx
10
)e−
(2n+1)2π2 t100 (83)
Plugging x = 5 in above expression will give us the temperature at midpoint of the rod.
7In hindsight, Fourier Cosine coefficients vanish.
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7 Coda
Well, I hope you enjoyed our little journey. At the end of the day, let me say something about life (Relax, I’m
not gonna preach you or whatnot). Life becomes more pleasant if you add every now and then some literature
(among other cool things) into your everyday rides in physics and mathematics. That’s why I would like to end
this piece with a quote from “Absalom! Absalom!” by the great William Faulkner. Have fun!
“You get born and you try this and you don’t know why only you keep on trying it and you are born at the
same time with a lot of other people, all mixed up with them, like trying to, having to, move your arms and legs
with strings only the same strings are hitched to all the other arms and legs and the others all trying and they
don’t know why either except that the strings are all in one another’s way like five or six people all trying to make
a rug on the same loom only each one wants to weave his own pattern into the rug; and it can’t matter, you know
that, or the Ones that set up the loom would have arranged things a little better, and yet it must matter because
you keep on trying or having to keep on trying and then all of a sudden it’s all over and all you have left is a
block of stone with scratches on it provided there was someone to remember to have the marble scratched and set
up or had time to, and it rains on it and then sun shines on it and after a while they don’t even remember the
name and what the scratches were trying to tell, and it doesn’t matter. And so maybe if you could go to someone,
the stranger the better, and give them something-a scrap of paper-something, anything, it not to mean anything
in itself and them not even to read it or keep it, not even bother to throw it away or destroy it, at least it would
be something just because it would have happened, be remembered even if only from passing from one hand to
another, one mind to another, and it would be at least a scratch, something, something that might make a mark
on something that was once for the reason that it can die someday, while the block of stone can’t be is because it
never can become was because it can’t ever die or perish...”
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