MATH3121 – Mathematical Methods and
PDEs
Topic 5: Generalised Fourier Series
Dr Chris Angstmann
School of Mathematics and Statistics
The University of New South Wales
Sydney, Australia
1
History of the Fourier Series
At the beginning of the 19th century the French mathematician
Jean Baptiste Joseph Fourier wrote Treatise on the
Propagation of Heat in Solid Bodies. This work, and his latter
book Analytical Theory of Heat, had a profound impact on the
development of mathematics.
One of the Fourier’s many insightful realisations was that
arbitrary continuous functions can be expressed as an infinite
trigonometric series.
He used such series to write general solutions to the heat
equation.
2
Fourier Series
Fourier, somewhat informally, showed that we can represent an
arbitrary continuous function, f (x), as a convergent series in
the elementary trigonometric functions,
f (x) =
∞∑
k=0
ak cos(kx) + bk sin(kx).
We require a little more formalism before we can calculate the
coefficients, ak and bk .
Remember that this infinite sum is really a limit,
f (x) = lim
N→∞
N∑
k=0
ak cos(kx) + bk sin(kx).
Naturally convergence issues need to be worked out with such
a limit.
3
Generalised Fourier Series
We will look at the more general problem of series
representations of functions.
Suppose that {φn(x)}∞n=0 is a set of orthogonal functions with
respect to a weight function w(x) on the interval (a,b).
Let f (x) be an arbitrary function defined on (a,b).
Can we represent f as a series in the set of orthogonal
functions, i.e.
f (x) =
∞∑
k=0
ckφk (x),
and if so how do we calculate the coefficients ck?
4
Inner Product
We denote by
〈f ,g〉 =
∫ b
a
f (x)g(x)w(x)dx
the inner product of f (x) and g(x) with respect to the weight
function w(x) on the interval (a,b).
Hence, if f and g are orthogonal, then 〈f ,g〉 = 0. Also, the
eigenfunctions of the Sturm-Liouville problem 〈φm, φn〉 = 0 for
n 6= m by definition.
5
An aside about Inner Products
I don’t expect you to recall this for an exam
In general an inner product must obey four properties.
1. 〈f + h,g〉 = 〈f ,g〉+ 〈h,g〉
2. 〈αf ,g〉 = α〈f ,g〉
3. 〈f ,g〉 = 〈g, f 〉
4. 〈f , f 〉 ≥ 0 with 〈f , f 〉 = 0 iff f = 0.
It is easy to see that the integral inner product defined above
satisfies all these properties, provide that w(x) > 0 for
x ∈ (a,b).
6
A Norm
Generally, w(x) > 0 on (a,b). Hence the inner product of f (x)
with itself, 〈f , f 〉 is non-negative (property 4). This means that
we can use the inner product to define how “long" a function is.
For w(x) > 0 on (a,b), we define the norm of f to be
‖f‖ =
√
〈f , f 〉 =
(∫ b
a
[f (x)]2w(x)dx
)1/2
7
Finding Coefficients
f (x) =
∞∑
k=0
ckφk (x).
Taking the inner product of both sides with φj(x) to get
〈f , φj〉 =
〈 ∞∑
k=0
ckφk , φj
〉
=
∞∑
k=0
ck 〈φk , φj〉
= cj〈φj , φj〉 since 〈φk , φj〉 = 0 when j 6= k
Therefore,
cj =
〈f , φj〉
‖φj‖2
.
This formula takes an even simpler form if each of the functions
φj has unit norm, i.e. ‖φj‖ = 1 for all j .
8
Orthonormal Set
An set of functions {ψk} is called an orthonormal set if it is an
orthogonal set and each of the ψk has unit norm, i.e. ‖ψk‖ = 1
for all k .
We can always normalise a given set of orthogonal functions
{φk}. To do this we set
ψk =
φk
‖φk‖ .
Then {ψk} is an orthonormal set.
‖ψk‖ =
√
〈ψk , ψk 〉
=
√〈
φk
‖φk‖ ,
φk
‖φk‖
〉
=
√〈φk , φk 〉
‖φk‖ = 1
9
Fourier Coefficients
The coefficients cn in the generalised Fourier series with
respect to the orthonormal set, {ψk},
f (x) =
∞∑
k=0
ckψk (x),
are given by ck = 〈f , ψk 〉.
Note: We have implicitly assumed that it is possible to choose
the coefficients ck so that the sum
N∑
k=0
ckφk (x)
converges to f (x) as N →∞. Later we will define precisely
what is meant by convergence and under what conditions
Fourier series converge.
10
Fourier-Lengendre Series
The Legendre polynomials {Pn(x)}∞n=0 are orthogonal
polynomials with respect to the weight function w(x) = 1 on the
interval (−1,1).
The norm squared of Pn is,
‖Pn‖2 =
∫ 1
−1
[Pn(x)]2dx =
2
2n + 1
.
11
Fourier-Legendre Series
The Fourier-Legendre series for a function f (x) defined on the
interval (−1,1) is
f (x) =
∞∑
n=0
cnPn(x) cn =
〈f ,Pn〉
‖Pn‖2
Since ‖Pn‖2 = 22n + 1, we have
cn =
2n + 1
2
∫ 1
−1
f (x)Pn(x)dx
12
Fourier-Legendre Example
Find the first three terms in the Fourier-Legendre series of the
function
f (x) =
{
0, −1 ≤ x ≤ 0,
x , 0 ≤ x ≤ 1.
P0(x) = 1, P1(x) = x , P2(x) = 12(3x
2 − 1)
13
Fourier-Legendre Example
Sketch 14P0(x) +
1
2P1(x) +
5
16P2(x).
Compare with the function f (x) from the last slide.
Then find the next term in the series and sketch the sum of the
first 4 terms.
Observe how the sum converges to f (x).
14
Fourier Sine Series
The Fourier Sine series uses the orthogonal set {sin nx}∞n=1 on
0 ≤ x ≤ pi.
Here the norm squared is,
‖ sin nx‖2 =
∫ pi
0
sin2(nx)dx
=
1
2
∫ pi
0
(1− cos(2nx))dx
=
1
2
[
x − sin(2nx)
2n
]pi
0
=
pi
2
15
Fourier Sine Series
The Fourier Sine series for f (x) on 0 ≤ x ≤ pi is
f (x) =
∞∑
n=1
cn sin nx
where
cn =
2
pi
∫ pi
0
f (x) sin nxdx .
16
Fourier Sine Series Example
Find the first six terms in the Fourier-Sin series of the function
f (x) = cos(x), 0 ≤ x ≤ pi.
17
Fourier Bessel Series
For any fixed m, the functions{
Jm
(
µ
(m)
n x
a
)}∞
n=1
are orthogonal w.r.t the weight function w(x) = x on 0 < x < a.
Here µ(m)n is the nth zero of Jm(µ).
The Fourier Bessel series for a function f (x) defined on (0,a) is
f (x) =
∞∑
n=1
cnJm
(
µ
(m)
n x
a
)
where
cn =
∫ a
0 xf (x)Jm
(
µ
(m)
n x
a
)
dx
∫ a
0 x
[
Jm
(
µ
(m)
n x
a
)]2
dx
18
Fourier Bessel Series
The integral in the denominator (i.e. ‖Jm‖2) can be evaluated
(with some difficulty)
∫ a
0
x
[
Jm
(
µ
(m)
n x
a
)]2
dx =
a2
2
[
Jm+1(µ
(m)
n )
]2
So,
cn =
2
a2
∫ a
0 xf (x)Jm
(
µ
(m)
n x
a
)
dx[
Jm+1(µ
(m)
n )
]2
19
Fourier Bessel Series Example
Find the first three terms in the Fourier-Bessel series of the
function
f (x) = sin(pix), 0 ≤ x ≤ 1,
using m = 1 and m = 2.
20
A Second Example
Find the Fourier-Bessel series for f (r) = 1− r2, 0 ≤ r ≤ 1 with
respect to the orthogonal set
{
J0
(
µ
(0)
n r
)}
.
21
Chladni Plate
At the beginning of the 19th century Ernest Chladni studied the
vibration of plates. In his simple experiments he vibrated a
fixed, circular plate with a violin bow and then sprinkling fine
sand across it to show the various nodal lines and patterns.
Stone, Elementary Lessons on Sound (1897)
https://www.youtube.com/watch?v=tFAcYruShow
22
A Vibrating Circular Membrane
Rather then concern ourselves with complicated geometries at
the moment, we will just consider a vibrating circular
membrane.
http://www.youtube.com/watch?v=v4ELxKKT5Rw
23
A Vibrating Circular Membrane
Consider a vibrating circular membrane. The displacement
u(r , θ, t) from equilibrium satisfies the wave equation
∇2u = 1
c2
∂2u
∂t2
, 0 ≤ r ≤ a,
with boundary condition u(a, θ, t) = 0 for some initial conditions.
In polar coordinates, the wave equation becomes
1
r
∂
∂r
(
r
∂u
∂r
)
+
1
r2
∂2u
∂θ2
=
1
c2
∂2u
∂t2
24
Example
Step 1 Put u(r , θ, t) = R(r)Θ(θ)T (t).
Step 2 Substitute into the PDE,
ΘT
r
d
dr
(
r
dR
dt
)
+
RT
r2
d2Θ
dθ2
=
RΘ
c2
d2T
dt2
.
Divide throughout by RΘT ,
1
rR
d
dr
(
r
dR
dr
)
+
1
r2Θ
d2Θ
dθ2
=
1
c2T
d2T
dt2
= −λ.
25
Example
The left-hand-side is a function of r and θ, but the
right-hand-side is a function of t only, so both functions equal a
const, as they are independent variables.
1
rR
d
dr
(
r
dR
dr
)
+
1
r2Θ
d2Θ
dθ2
= −λ
r
R
d
dr
(
r
dR
dr
)
+ λr2 = − 1
Θ
d2Θ
dθ2
= −β
This now give us three ODEs for T ,R, and Θ.
26
Example
Step 3. Consider the problem for Θ.
Θ′′ − βΘ = 0
If θ increases by 2pi we return to the same point on the
membrane. Thus Θ(θ) must have a period of 2pi if the solution
is to be single valued.
This means that β = −m2 where m is an integer, since the
solution in this case is
Θm(θ) = Am cos mθ + Bm sin mθ, m = 0,1,2, . . . .
27
Example
Consider now the equation for R(r),
d
dr
(
r
dR
dr
)
+
(
λr − m
2
r
)
R = 0
This is a Sturm-Liouville type problem, with p(r) = r . The
boundary condition u(a, θ, t) = 0 implies that R(a) = 0. But, is
there a boundary condition at r = 0?
Since p(0) = 0, the equation is singular. So the boundary
conditions R and R′ finite as r → 0+ are sufficient.
These must hold from the physics of the problem.
28
Example
Case 1 If λ > 0, put λ = κ2. Let ρ = κr . Then
dR
dr
=
dR
dρ
dρ
dr
= κ
dR
dρ
r
dR
dr
= ρ
dR
dρ
Substitute these we have
ρ
d
dρ
(
ρ
dR
dρ
)
+ (ρ2 −m2)R = 0
which is a Bessels equation of order m, with general solution
R(ρ) = AJm(ρ) + BYm(ρ)
R(r) = AJm(κr) + BYm(κr)
29
Example
Since |Ym(ρ)| → ∞ as ρ→ 0, we choose B = 0 to satisfy the
condition that R and R′ are finite as ρ→ 0.
The boundary condition R(a) then implies
AJm(κa) = 0.
A = 0 is the trivial solution, so κa = zero of Jm.
Let µ(m)n be the nth root of Jm(µ) = 0. So
κna = µ
(m)
n , or κn =
µ
(m)
n
a
, n = 1,2,3, . . .
30
Example
Hence the positive eigenvalues are
λn = κ
2
n =
(
µ
(m)
n
a
)2
with eigenfunctions
Rn(r) = Jm
(
µ
(m)
n
a
r
)
, n = 1,2,3, . . .
31
Example
Case 2 If λ = 0, the ODE for R reduces to
r
d
dr
(
r
dR
dr
)
= m2R
or r2R′′ + rR′ −m2R = 0. This is a Euler equation with general
solution
R(r) =
{
Arm + Br−m, m 6= 0
A + B ln r , m = 0
Since R is finite at r = 0, B = 0 in both cases, and therefore
R(r) = Arm for all m.
Now R(a) = 0, so A = 0 giving the trivial solution, so λ = 0 is
not an eigenvalue.
32
Example
Case 3 If λ < 0, let λ = −τ2, the differential equation becomes
r
d
dr
(
r
dR
dr
)
+ (r2τ2 −m2)R = 0.
As before, let ρ = τ r and the equation becomes
ρ
d
dρ
(
ρ
dR
dρ
)
+ (−ρ2 −m2)R = 0.
This is the modified Bessel equation, and it has the general
solution
R(r) = A′Jm(iρ) + B′Ym(iρ) = AIm(ρ) + BKm(ρ),
where Im and Km are modified Bessel functions.
33
Example
As ρ→ 0, Km is unbounded and so B = 0. The function Im(ρ) is
never zero for ρ > 0 and hence the boundary condition
R(a) = 0 implies A = 0, so again the trivial solution results and
hence there are no negative eigenvalues.
34
Example
It only remain now to solve for T (t), which satisfies
T ′′ = −λc2T
Putting λ = κ2n and solving we obtain
Tn(t) = αn sinκnct + βn cosκnct .
So a separable solution for the vibrating membrane is
umn(r , θ, t) =
(Am cos mθ + Bm sin mθ)(αn sinκnct + βn cosκnct)Jm
(
µ
(m)
n
a
)
where µ(m)n is the nth root of Jm, κn = µ
(m)
n /a.
35
Example
The next step is to construct a linear combination of separable
solutions which satisfy the initial conditions. Let us simplify the
problem by supposing that the initial conditions are
independent of θ, i.e.
u(r , θ, 0) = f (r),
∂u
∂t
(r , θ, 0) = g(r)
Hence we need only look for separable solutions that are
independent of θ, (i.e. put m = 0 in the expression for umn).
So using the principle of superposition we get
u(r , t) =
∞∑
n=1
(αn sinκnct + βn cosκnct)J0
(
µ
(0)
n r
a
)
where κn = µ
(0)
n /a.
36
Example
Now applying the initial conditions we obtain:
f (r) =
∞∑
n=1
βnJ0
(
µ
(0)
n r
a
)
and
g(r) =
∞∑
n=1
cκnαnJ0
(
µ
(0)
n r
a
)
, for 0 ≤ r ≤ a.
We must expand f (r) and g(r) into a Fourier-Bessel series to
find the constants αn and βn.
We do this by using the orthogonality of the Bessel functions.
We saw that (when m = 0),∫ a
0
rJ0
(
µ
(0)
n r
a
)
J0
(
µ
(0)
p r
a
)
dr = 0, n 6= p.
37
Example
Consider
f (r) =
∞∑
n=1
βnJ0
(
µ
(0)
n r
a
)
, 0 ≤ r ≤ a.
Multiply both sides with rJ0
(
µ
(0)
p r
a
)
and integrate w.r.t r from 0
to a.∫ a
0
rf (r)J0
(
µ
(0)
p r
a
)
dr =
∞∑
n=1
βn
∫ a
0
rJ0
(
µ
(0)
n r
a
)
J0
(
µ
(0)
p r
a
)
dr
= βp
∫ a
0
r
[
J0
(
µ
(0)
p r
a
)]2
dr
Thus the coefficients βn are given by
βn =
∫ a
0
rf (r)J0
(
µ
(0)
p r
a
)
dr
/∫ a
0
r
[
J0
(
µ
(0)
p r
a
)]2
dr
38
Convergence of Generalised Fourier Series
A motivating example
Consider the function f (x) = x for −pi < x ≤ pi.
The Fourier series representation of this function is,
f (x) = 2
∞∑
k=1
(−1)k+1
k
sin(kx).
But what happens when the sum is not infinite? We can
consider the partial sum,
Tn(x) = 2
n∑
k=1
(−1)k+1
k
sin(kx).
How well does this approximate the function f (x)?
39
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
f
T10
40
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
f
T100
41
Approximation in the Mean
Suppose that {φn(x)} is a given set of orthonormal functions
and that we want to approximate some function f (x) by
Tn(x) =
n∑
k=0
akφk (x), a ≤ x ≤ b,
where n is a fixed integer, and ak can be any real numbers.
Let the weighted mean square error E2n be defined by
E2n = ‖f − Tn‖2 = 〈f − Tn, f − Tn〉
= 〈f , f 〉 − 2〈f ,Tn〉+ 〈Tn,Tn〉.
42
Approximation in the Mean
Now
〈f ,Tn〉 = 〈f ,
n∑
k=0
akφk 〉 =
n∑
k=0
ak 〈f , φk 〉.
and
〈Tn,Tn〉 =
n∑
k=0
n∑
j=0
akaj〈φk , φj〉 =
n∑
k=0
a2k .
Hence
E2n = ‖f‖2 − 2
n∑
k=0
ak 〈f , φk 〉+
n∑
k=0
a2k
= ‖f‖2 +
n∑
k=0
[ak − 〈f , φk 〉]2 −
n∑
k=0
|〈f , φk 〉|2
43
Approximation in the Mean
So the weighted mean square error, E2n , is minimized by
choosing ak = 〈f , φk 〉.
Theorem 1
The coefficients that make E2n a minimum are the Fourier
coefficients ak = 〈f , φk 〉 for k = 0,1,2, . . . ,n.
Let Sn(x) =
∑n
k=0〈f , φk 〉φk (x) be the truncated (partial Fourier
series). Then
‖f (x)− Sn(x)‖2 = ‖f‖2 −
n∑
k=0
|〈f , φk 〉|2 = E2n ≥ 0.
44
Approximation in the Mean
Therefore,
n∑
k=0
|〈f , φk 〉|2 ≤ ‖f‖2
Thus,
∑∞
k=0 |〈f , φk 〉|2 exists, as the series has an upper bound,
even though the Fourier series for f may not converge, and
∞∑
k=0
|〈f , φk 〉|2 ≤ ‖f‖2 Bessel’s inequality
It also implies that lim
k→∞
|〈f , φk 〉| = 0 as the series converges.
45
Approximation in the Mean
Furthermore, since
lim
n→∞ ‖f − Sn‖
2 = ‖f‖2 −
∞∑
n=0
|〈f , φn〉|2
Then
lim
n→∞ ‖f − Sn‖
2 = 0 if and only if
‖f‖2 =
∞∑
n=0
|〈f , φn〉|2 Parseval’s Identity
We say a Fourier series converges in the mean to f (x) if and
only if Parseval’s Identity holds.
46
Convergence of Fourier series
What is meant by convergence?
I We could mean that
lim
n→∞ |f (x)− Sn(x)| = 0 for every x ∈ [a,b].
This is called point-wise convergence.
I Or, we could say that
lim
n→∞ maxx∈[a,b]
|f (x)− Sn(x)| = 0,
which is known as uniform convergence.
I Or we could mean that
lim
n→∞ ‖f (x)− Sn(x)‖ = 0
which is called convergence in the mean. Fourier series
usually converge in this sense.
47
Piecewise continuous functions
A function f (x) is piecewise continuous on [a,b] if it is
continuous everywhere in [a,b] except at a finite number of
points where there are simple discontinuities, i.e. the left hand
and righ hand limits exist.
48
Piecewise continuous functions
Denote the class of all functions that are piecewise continuous
on the interval [a,b] as Cp[a,b].
In discussing the convergence of Fourier series we will restrict
ourselves to piecewise continuous functions.
49
Piecewise continuous functions
Let w(x) be a continuous and positive function for a < x < b
such that ∫ b
a
w(x)dx exists.
It can be shown that if F (x) ∈ Cp[a,b] then∫ b
a
F (x)w(x) also exists.
Thus for f ,g ∈ Cp[a,b] then 〈f ,g〉 and ‖f‖ and ‖g‖ exist if the
inner product is defined with respect to the weight function
w(x) on [a,b].
50
Convergence in the Mean
Suppose {Sn}, n ≥ 0, is a sequence of functions from Cp[a,b]
and the function S(x) is also in Cp[a,b]. Then the sequence
{Sn(x)} converges in the mean to S(x) with respect to the
weight function w(x) on [a,b] if
lim
n→∞ ‖Sn − S‖ = 0
i.e.
lim
n→∞
∫ b
a
(Sn(x)− S(x))2w(x)dx = 0
51
Complete (or closed) orthonormal sets
An orthonormal set {φn} is said to be complete (or closed) in
the function space Cp[a,b] if for every f ∈ Cp[a,b] the Fourier
series for f converges in the mean to f .
That is, {φn} is complete if for all f ∈ Cp[a,b]
lim
n→∞ ‖f − Sn‖ = 0, where Sn =
n∑
k=0
〈f , φk 〉φk .
Thus, from the previous section, an orthonormal set is
complete if and only if
∞∑
k=0
〈f , φk 〉2 = ‖f‖2,
i.e. Parseval’s Identity holds for every f ∈ Cp[a,b].
52
Complete (or closed) orthonormal sets
Theorem 2
If the set {φn} is complete in Cp[a,b] and if f is orthogonal to
each φn the f must be zero except at possibly a finite number of
points in [a,b].
Proof Now 〈f , φn〉 = 0 for all n, i.e. every Fourier coefficient of f
is zero. Since {φn} is complete and the Fourier series for f
converges in the mean and Parseval’s Identity holds. So
‖f‖2 =
∞∑
n=0
〈f , φn〉2 = 0
i.e. ∫ b
a
(f (x))2w(x)dx = 0
and since w(x) > 0 on [a,b] and f 2 is non-negative, then
f (x) = 0 except at maybe some finite number of isolated points.
53
Complete (or closed) orthonormal sets
There is no “area” under single points.
54
Important consequences
I This implies that if one or more members (not all zero) are
deleted from the complete orthogonal set {φn}, then the
set is no longer complete since the deleted members are
orthogonal to each member of the new set.
I Hence we can define a complete orthogonal set as one
where it is impossible to add another non-zero (except
perhaps at a finite number of points) function which is
orthogonal to each member of the set.
I This also provides us with a way of testing whether or not a
set is not complete.
55
Example
Take φn(x) = sin nx , n = 2,3,4, . . . on [0, pi]. This set is
clearly orthogonal on [0, pi] with weight w(x) = 1 but is not
complete as sin x is not included, and this is orthogonal to each
sin nx , for n = 2,3,4, . . ..
However it can be shown that sin nx , n = 1,2,3, . . . is a
complete set on [0, pi].
Take φn(x) = sin nx , for n = 1,2,3, . . . on [−pi, pi]. This set is
simply orthogonal on [−pi, pi], but is not complete as you need
to include cos kx , k = 0,1,2, . . . on this interval.
56
Examples of complete orthogonal sets
A set of polynomials φn is called simple if each φn has degree n
for n = 0,1,2, . . .. Any set of simple polynomials {φn}∞n=0 that is
orthogonal on a finite interval w.r.t a weight function w(x) is
complete in the space Cp[a,b].
Example:
The Legendre polynomials {Pn(x)} on [−1,1] is a complete
orthogonal set (w(x) = 1). Thus the Fourier series for any
piecewise continuous function defined on [−1,1] will converge
in the mean to f (x).
Example:
The set of all eigenfunctions of the Sturm-Liouville equation
with separated boundary conditions form a complete
orthogonal set. Any piecewise continuous function defined on
[a,b] can be represented by a Fourier series of these
orthogonal functions which will converge in the mean to the
function.
57
Conduction of heat along a slender rod
Consider the conduction of heat along a slender rod with
insulated edges.
Suppose the rod is so thin that the temperature is uniform over
any cross-section. So denote the temperature by u(x , t), where
x is the distance along the rod, and t is time.
It can be shown that the heat satisfies the 1D Heat Equation
∂u
∂t
= κ
∂2u
∂x2
, (κ is the thermal diffusivity)
58
Typical Boundary Equations
Either the temperature u or its derivative ∂u/∂x is given at the
ends of the rod (Dirichlet or Neumann conditions resp.)
The quantity ∂u/∂x is proportional to the rate at which heat
passes through the cross-section of the rod at x and so by
specifying ∂u/∂x at the ends, we are, in effect, specifying the
heat flux at the ends of the rod. E.g. if the rod is also insulated
at x = ` then the heat flux is zero at that end and the
corresponding boundary condition is
∂u
∂x
∣∣∣∣
x=`
= 0.
In addition to the boundary condition, the initial condition, the
temperature at t = 0, u(x ,0) is usually given.
59
Non-dimensional form
Let
x∗ =
x
`
, where ` is the length of the rod,
and
u∗ =
u
u0
,
where u0 is a typical temperature which appears in the
specification of the problem, e.g. the max of average
temperature as specified by u(x ,0).
On substituting these into the PDE we obtain
∂u∗
∂t
=
κ
`2
∂2u
∂x∗2
.
60
We can eliminate κ and ` from the problem by choosing
t∗ =
tκ
`2
Hence the non-dimensional heat equation is
∂u∗
∂t∗
=
∂2u∗
∂x∗2
, u∗ = u∗(x∗, t∗)
subject to the boundary conditions where either u∗ or ∂u∗/∂x∗
are specified at x∗ = 0 or x∗ = 1, and with initial condition
u∗(x∗,0) = f (x∗), 0 < x∗ < 1.
From now on we use only dimensionless variables (so drop the
asterisks).
61
Example: Homogeneous Boundary Conditions
Take the equation,
∂u
∂t
=
∂2u
∂x2
on 0 < x < 1, t > 0
where u(0, t) = u(1, t) = 0 and u(x ,0) = f (x), 0 < x < 1.
Let u(x , t) = X (x)T (t) etc, and the solution (as an exercise) is
u(x , t) =
∞∑
n=1
Bne−n
2pi2t sin(npix)
where
Bn = 2
∫ 1
0
f (x) sin(npix)dx
62
Example Non-homogeneous boundary conditions
As previous example, but suppose that we have specified
temperatures at the boundary, but they are non-zero, i.e.
u(0, t) = a and u(1, t) = b.
This now means that separation of variables does not work.
Why?
Suppose u1 and u2 are both solutions, then
u1(0, t) = u2(0, t) = a.
But, u1 + u2 should also be a solution, however
u1(0, t) + u2(0, t) 6= a unless a = 0.
63
Hence the Principle of Superposition does not work if you have
non-homogeneous boundary conditions.
The separation of variables method can only be applied if
boundary conditions are homogeneous.
However, we can solve the problem if we can convert it to a
problem with homogeneous boundary conditions.
Let u(x , t) = φ(x) + v(x , t) where φ(x) is the steady state
temperature distribution such that φ(0) = a and φ(1) = b. This
means that v(x , t) satisfies the PDE and has homogeneous
boundary conditions.
64
Steady state heat flow - Laplace’s equation
We have seen that in a 2D region the temperature, u, satisfies
the 2D heat equation
∇2u = 1
κ
∂u
∂t
.
In order to find the steady state temperature, assume the
temperature, u, is independent of time throughout the region.
We must solve
∇2u = 0
in the region where the temperature on the boundary is given -
this is a Dirichlet Problem.
65
Laplace’s equation in a rectangular region
Solve
∂2u
∂x2
+
∂2u
∂y2
= 0
for 0 < x < a and 0 < y < b.
The boundary
conditions are
u = 0 on x = 0, a 0 < y < b
u = 0 on y = 0 0 < x < a
u = f (x) on y = b 0 < x < a
66
Let u(x , y) = X (x)Y (y) and hence X ′′Y + XY ′′ = 0 or
X ′′
X
= −Y
′′
Y
= −λ.
Thus X ′′ + λX = 0 and Y ′′ − λY = 0. The boundary conditions
may also be separated.
u(0, y) = 0 and u(a, y) = 0⇒ X (0)Y (y) = 0 and X (a)Y (y) = 0.
If Y (y) = 0 then u ≡ 0 which doesn’t satisfy u(x ,b) = f (x).
Hence X (0) = 0 and X (a) = 0.
Similarly, u(x ,0) = 0⇒ X (x)Y (0) = 0⇒ Y (0) = 0.
67
It is best to look at the X equation first as it has two
homogeneous boundary conditions. The problem for X (x) is
X ′′ + λX = 0, where X (0) = X (a) = 0.
There is no non-trivial solutions for λ ≤ 0 (Excercise). For
λ > 0, the solution to the ODE is
X (x) = c1 cos
√
λx + c2 sin
√
λx
Boundary conditions: X (0) = 0⇒ c1 = 0 and
X (a) = 0⇒ c2 sin
√
λa = 0⇒ √λa = npi, n = 1,2,3, . . ..
68
Therefore the eigenvalues are
λn =
n2pi2
a2
and the corresponding eigenfunctions are
Xn(x) = Cn sin
npix
a
.
So Yn(y) satisfies
Y ′′n −
n2pi2
a2
Yn = 0, Yn(0) = 0.
Hence
Yn(y) = An cosh
npiy
a
+ Bn sinh
npiy
a
Since Yn(0) = 0, we have An = 0 and hence
un(x , y) = Dn sin
npix
a
sinh
npiy
a
69
Let
u(x , y) =
∞∑
n=1
Dn sin
npix
a
sinh
npiy
a
Now
u(x ,b) = f (x) =
∞∑
n=1
Dn sin
npix
a
sinh
npib
a
, 0 < x < a
i.e.
f (x) =
∞∑
n=1
bn sin
npix
a
.
This is a Fourier Sine series for f (x) and so
bn =
2
a
∫ a
0
f (x) sin
npix
a
,
from which Dn can be easily computed.
70
Laplace’s Equation in a Circular Disc
We want to find u(r , θ)
satisfying
∇2u = 1
r
∂
∂r
(
r
∂u
∂r
)
+
1
r2
∂2u
∂θ2
= 0
u(a, θ) = f (θ).
71
Laplace’s Equation in a Circular Disc
Let u(r , θ) = R(r)Θ(θ) and hence
1
r
d
dr
(
r
dR
dr
)
Θ +
R
r2
d2Θ
dθ2
= 0.
Multiplying throughout by r2/(RΘ) and rearranging we have
r
R
d
dr
(
r
dR
dr
)
= − 1
Θ
d2Θ
dθ2
= λ, (const).
The ODE for Θ is
Θ′′ + λΘ = 0.
There are implicit boundary conditions on Θ. Since we are
looking for 2pi periodic solutions in θ, λ must = n2, n is an
integer. Hence,
Θn(θ) = An cos nθ + Bn sin nθ, n = 0,1,2, . . .
72
Laplace’s Equation in a Circular Disc
The ODE for R is then
r2R′′ + rR′ − n2R = 0
which is an ODE of the Euler-Cauchy type. This can be solved
by changing the variable, r = es, leading to an ODE with
constant coefficients.
The solution is
Rn(r) =
{
E1 + D1 ln r , n = 0
E2rn + D2r−n, n ≥ 1.
The solution for u must be finite on the disc. Both ln r and r−n
are unbounded when r = 0, so D1 = D2 = 0 for all n ≥ 0. Thus,
Rn(r) = Ern, n = 0,1,2, . . .
Combining the solutions and using superposition:
u(r , θ) =
∞∑
n=0
rn(An cos nθ + Bn sin nθ)
73
Laplace’s Equation in a Circular Disc
Or
u(r , θ) =
1
2
A0 +
∞∑
n=1
rn(An cos nθ + Bn sin nθ)
Now since u(a, θ) = f (θ), we have
f (θ) =
1
2
a0 +
∞∑
n=1
an(An cos nθ + Bn sin nθ)
Thus anAn and anBn are the cosine and sine coefficients in the
general trigonometric Fourier series for f (θ). Hence,
An =
1
pian
∫ pi
−pi
f (θ) cos nθdθ, n = 0,1,2, . . .
and
Bn =
1
pian
∫ pi
−pi
f (θ) sin nθdθ, n = 1,2,3, . . .
74
Laplace’s Equation in a Sphere
Laplace’s equation in spherical coordinates is
∇2u = 1
r2
∂
∂r
(
r2
∂u
∂r
)
+
1
r2 sin2 θ
∂2u
∂φ2
+
1
r2 sin θ
∂
∂θ
(
sin θ
∂u
∂θ
)
= 0
and let us solve this subject to
u(a, θ, φ) =
{
u1 (const), 0 ≤ θ ≤ pi/2
u2 (const), pi2 < θ ≤ pi
75
Laplace’s Equation in a Sphere
Now since the boundary conditions are independent of φ we
can assume u = u(r , θ) only. Laplace’s equation therefore
becomes
∇2u = 1
r2
∂
∂r
(
r2
∂u
∂r
)
+
1
r2 sin2 θ
∂
∂θ
(
sin θ
∂u
∂θ
)
= 0
Let u(r , θ) = R(r)Θ(θ) then
1
R
d
dr
(
r2
dR
dr
)
= − 1
Θ sin θ
d
dθ
(
sin θ
dΘ
dθ
)
= λ, const
The ODE for Θ is
1
sin θ
d
dθ
(
sin θ
dΘ
dθ
)
+ λΘ = 0
76
Laplace’s Equation in a Sphere
Now let λ = s(s + 1) (s is a new const) and rearrange to
d2Θ
dθ2
+
cos θ
sin θ
dΘ
dθ
+ s(s + 1)Θ = 0.
Let µ = cos θ, then
dΘ
dθ
=
dΘ
dµ
dµ
dθ
= − sin θdΘ
dµ
d2Θ
dθ2
= sin2 θ
d2Θ
dµ2
− cos θdΘ
dµ
= (1− µ2)d
2Θ
dµ2
− µdΘ
dµ
77
Laplace’s Equation in a Sphere
Substituting,
(1− µ2)d
2Θ
dµ2
− µdΘ
dµ
− µdΘ
dµ
+ s(s + 1)Θ = 0,
(1− µ2)d
2Θ
dµ2
− 2µdΘ
dµ
+ s(s + 1)Θ = 0,
which is Legendre’s Equation. This has finite solutions for
−1 ≤ µ ≤ 1 (or 0 ≤ θ ≤ pi) only when s = n, n = 0,1,2, . . ..
The finite solutions are the Legendre Polynomials:
Pn(µ) = Pn(cos θ), n = 0,1,2, . . .,i.e.
Θn(θ) = Pn(cos θ), n = 0,1,2, . . .
78
Laplace’s Equation in a Sphere
The ODE for R becomes
r2R′′n + 2rR
′
n − n(n + 1)Rn = 0, n = 0,1,2, . . .
This is also an ODE of the Euler-Cauchy type and has a
general solution
R(r) = Arn +
B
rn+1
but if u is finite at r = 0 the Bn = 0.
Thus the solutions for u are un(r , θ) = AnrnPn(cos θ).
So
u(r , θ) =
∞∑
n=0
AnrnPn(cos θ).
79
Laplace’s Equation in a Sphere
To satisfy the boundary condition on the surface of the sphere
we require
u(a, θ) = f (θ) =
∞∑
n=0
AnanPn(cos θ)
or
f (cos−1 µ) =
∞∑
n=0
AnanPn(µ), µ = cos θ.
This is a Fourier-Legendre series and using
‖Pn‖2 =
∫ 1
−1[Pn(x)]
2dx = 2/(2n + 1),
An =
2n + 2
2an
∫ 1
−1
f (cos−1 µ)Pn(µ)dµ
or let µ = cos θ,dµ = − sin θdθ
An =
2n + 2
2an
∫ pi
0
f (θ)Pn(cos θ) sin θdθ
80
Laplace’s Equation in a Sphere
In this case,
f (θ) =
{
u1 (const), 0 ≤ θ ≤ pi/2
u2 (const), pi2 < θ ≤ pi
So
An =
2n + 1
2an
[∫ 1
0
u1Pn(µ)dµ+
∫ 0
−1
u2Pn(µ)dµ
]
With P0(µ) = 1,P1(µ) = µ, we have
A0 =
1
2
(u1 + u2)
and
A1 =
3
2a
[
u1
∫ 1
0
µdµ+
∫ 0
−1
µdµ
]
=
3
4a
(u1 − u2)
81