DIFFERENTIAL EQUATIONS & SERIES
LECTURE 1
INTRODUCTION TO LAPLACE TRANSFORMS
Motivation
The Laplace Transform is an important mathematical device for analysing many di↵erential equations
used to model engineering systems.
Consider the following system, a basis for modelling mechanical vibrations.
k
↵
M
x (t)
(t)
Figure 1
The block of mass M is subject to an external force (which may vary with time t), and is attached
to a spring and a damper with constants k and ↵.
A simple model for the dynamics of the block (based on Newton’s second law) is the linear ordinary
di↵erential equation (ODE)
Mx¨ + ↵x˙ + kx = (1)
We learnt how to solve this type of ODE in Stage 1 Engineering Maths (ENG1001).
The solution x(t), giving the position of the block, can be expressed in the form
x(t) = xc(t) + xp(t)
The complementary function xc(t) is the general solution to the homogeneous equation ( = 0)
Mx¨ + ↵x˙ + kx = 0
and is found by looking for solutions in the form xc(t) = emt.
The particular solution xp(t) is any solution of equation (1).
In Stage 1 we learnt how to find xp(t) by generalising the forcing term (t).
For example, if (t) = 3t then we would look for solutions in the form xp(t) = at+ b.
By substituting this into equation (1) we could find the required values of a and b.
ENG2011
Di↵erential Equations & Series
– 1 – lecture 1
Laplace Transforms I
mass 线性
XGkemttY lt
代⼊中⼼北
拉普拉斯变换是分析许多⽤于⼯程系统建模的微分⽅程的重要数学⼯具。
考虑下⾯的系统，它是机械振动建模的基础。
质量为M的块体受外⼒φ(可随时间t变化)的作⽤，并附在常数为k和α的弹簧和阻尼器上。
给出物体位置的解x(t)可以表示为
⼀个简单的模型(基于⽜顿第⼆定律)是线性常微分⽅程(ODE)。
互补函数xc(t)是⻬次⽅程(φ = 0)的通解
This approach for finding particular solutions works provided the forcing function is ‘simple’.
However, in many practical applications the external forcing may be quite complicated.
Consider the examples in the following figure:

t t t
Figure 2
This shows forcing terms that are either discontinuous or do not change smoothly.
In these cases it is very hard to predict the correct form of particular solutions xp.
Laplace Transforms provide a means to solve ODEs involving these more complicated forcing terms,
and we will learn how to do this in later lectures (4 & 5).
Before beginning our study of Laplace Transforms we note that these transforms are very important
in Control Theory: Conceptually we can think of the system given in figure 1, or by equation (1),
as follows
input
(t)
output
x(t)linear system
We often want to choose the input (t) to control the system output x(t).
Laplace Transforms are an important tool for analysing and controlling dynamical responses.
ENG2011
Di↵erential Equations & Series
– 2 – lecture 1
Laplace Transforms I
当强迫函数φ是“简单”时，这种求特解的⽅法是有效的。然⽽，在许多实际应⽤中，外部强迫可能相当复杂。
考虑下图中的例⼦:
这表明强制项要么是不连续的，要么是不平稳变化的。
在这些情况下，很难预测特解xp的正确形式。
拉普拉斯变换提供了⼀种⽅法来解决涉及这些更复杂强迫项的ode，我们将在后⾯的讲座(4和5)中学习如何做到这⼀点。
在我们开始研究拉普拉斯变换之前，我们注意到这些变换在控制论中是⾮常重要的:从概念上讲，
我们可以将图1所示的系统，或通过⽅程(1)，考虑如下
Laplace Transforms: Definition
Given a function f(t) define
F (s) =
1Z
0
f(t) est dt (⇤)
Note the parameter s in this integral. As s changes so does the resulting value of the integral.
So the value of the integral, F , is a function of s.
F (s) is called the Laplace Transform of f(t), and we write
F (s) = L
h
f(t)
i
The operator L transforms the given function f(t) into another function F (s)
f (t) F (s) =
1Z
0
f (t) est dtL
Note that this only makes sense if the integral in definition (⇤) exists.
In other words, the area under the curve f(t) est, 0 6 t <1, must be finite.
This will usually restrict the values of s that are allowed.
For example if f(t) = t then we must take s > 0 (see figure 3 )
t est
area
A
A =
1Z
0
t est dt = F (s)
t
Figure 3
ENG2011
Di↵erential Equations & Series
– 3 – lecture 1
Laplace Transforms I
Let us determine, using definition (⇤), the Laplace Transform of the function f(t) = t.
We will use integration by parts.
L⇥ t ⇤ = 1Z
0
t|{z}
u
est|{z}
v0
dt
=
h
t|{z}
u

1
s e
st| {z }
v
i1
0

1Z
0
1|{z}
u0

1
s e
st| {z }
v
dt
When we evaluate the first term at the lower limit t = 0 we obviously get 0.
The first term is also 0 as t!1 (the upper limit), provided s > 0 (see graph in figure 3 ).
Therefore
L⇥ t ⇤ = 1Z
0
1|{z}
u0

1
s e
st| {z }
v
dt =
1Z
0
1
s e
st dt =
h
1
s2 e
st
i1
0
=
⇣
0 1s2
⌘
=
1
s2
So we have established that the Laplace Transform of the function f(t) = t is F (s) =
1
s2
.
That is
L⇥ t ⇤ = 1
s2
Using similar integration techniques we can determine from first principles, using definition (⇤), the
Laplace Transforms of other standard functions.
The table on the next page gives the Laplace Transforms F (s) of some standard functions f(t).
You are asked to derive some of these results in Exercises 1.
ENG2011
Di↵erential Equations & Series
– 4 – lecture 1
Laplace Transforms I
f (t) F (s)L
f(t) F (s)
1
1
s
9>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>;
s > 0
t
1
s2
tn
n = 1, 2, 3, . . .
n!
sn+1
sin(at)
a
s2 + a2
cos(at)
s
s2 + a2
sinh(at)
a
s2 a2
9>=>; s > |a|cosh(at) s
s2 a2
eat
1
s a s > a
H(t a) 1
s
eas s > 0, (a > 0)
Table of Transforms
(a is any constant)
In this course we will only consider Laplace Transforms in which s is a real number.
However the theory can be extended to include s as a complex number.
ENG2011
Di↵erential Equations & Series
– 5 – lecture 1
Laplace Transforms I
In the Table of Transforms there is a function that you may not have met before H(t a).
This is the Heaviside step function, named after the electrical engineer Oliver Heaviside (1850-1925).
This step function is defined by
H(t a) =
8<: 0 , t < a1 , t > a
The graph of H(t a) is
H
0 a t
1
We can think of H(t a) as a mathematical way of representing the e↵ect of a ‘switch’:
The switch is o↵ (H = 0) when t < a, and on (H = 1) when t > a.
To finish this lecture let us derive the final result in the Table of Transforms
L⇥H(t a)⇤ = 1Z
0
H(t a) est dt =
1Z
a
1 est dt
=
h
1
s e
st
i1
a
= 1
s
⇣
0 eas
⌘
=
1
s
eas
ENG2011
Di↵erential Equations & Series
– 6 – lecture 1
Laplace Transforms I
DIFFERENTIAL EQUATIONS & SERIES
LECTURE 2
FURTHER RESULTS ON LAPLACE TRANSFORMS
The Laplace Transforms of some standard functions are given in our Table of Transforms.
f(t) F (s)
1
1
s
9>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>;
s > 0
t
1
s2
tn
n = 1, 2, 3, . . .
n!
sn+1
sin(at)
a
s2 + a2
cos(at)
s
s2 + a2
sinh(at)
a
s2 a2
9>=>; s > |a|cosh(at) s
s2 a2
eat
1
s a s > a
H(t a) 1
s
eas s > 0, (a > 0)
Table of Transforms
(a is any constant)
We will need to determine the transforms of more complicated functions.
For example, how should we determine the Laplace Transform of f(t) = t2 e3t sin(4t) ?
In theory we could use the definition of a Laplace Transform and evaluate the integral
F (s) =
1Z
0
f(t) est dt =
1Z
0
t2 e3t sin(4t) est dt
However it is much easier to make use of some general rules for Laplace Transforms.1
These results will be important when we come to use Laplace Transforms to solve ODEs.
1compare with using di↵erentiation rules (product rule, chain rule . . .) instead of di↵erentiating from first principles
ENG2011
Di↵erential Equations & Series
– 1 – lecture notes 2
Laplace Transforms II
Some General Rules For Laplace Transforms
The following rules can be proved from the definition
F (s) =
1Z
0
f(t) est dt
(Proofs are outlined at the end of this lecture)
1 Linearity
L⇥af(t) + b g(t)⇤ = aL⇥f(t)⇤+ bL⇥g(t)⇤ a, b constants
2 First Shift Theorem
L⇥eatf(t)⇤ = F (s a)
3 Derivatives
L⇥f 0(t)⇤ = sF (s) f(0)
L⇥f 00(t)⇤ = s2F (s) sf(0) f 0(0)
4 Integral
L
Z t
0
f(u) du

=
1
s
F (s)
5 Multiplication by t
L⇥tf(t)⇤ = d
ds
F (s)
6 Second Shift Theorem
L⇥H(t a)f(t a)⇤ = easF (s)
ENG2011
Di↵erential Equations & Series
– 2 – lecture notes 2
Laplace Transforms II
Let’s do some examples of finding Laplace Transforms using these rules and our Table of Transforms :
(1) Find the Laplace Transform of 3t2 + 2et.
L⇥3t2 + 2et⇤ = 3L⇥t2⇤ + 2L⇥et⇤ using 1
= 3
2
s3
+ 2
1
s+ 1
from Table
(2) Find the Laplace Transform of t3e2t.
Use First Shift Theorem 2 : Take f(t) = t3 so that
L⇥t3e2t⇤ = L⇥f(t)e2t⇤ = F (s+ 2)
From Table
F (s) = L⇥f(t)⇤ = L⇥t3⇤ = 3!
s4
Therefore
L⇥t3e2t⇤ = F (s+ 2) = 3!
(s+ 2)4
(3) Find the Laplace Transform of e3t sin(4t).
Use First Shift Theorem 2 : Take f(t) = sin(4t) so that
L⇥e3t sin(4t)⇤ = L⇥e3tf(t)⇤ = F (s 3)
From Table
F (s) = L⇥f(t)⇤ = L⇥sin(4t)⇤ = 4
s2 + 16
Therefore
L⇥e3t sin(4t)⇤ = F (s 3) = 4
(s 3)2 + 16
(4) Use 3 to find the Laplace Transform of
d
dt
cos(t), and show that this is the same as the
transform of sin(t).
Take f(t) = cos(t) so that F (s) = L⇥f(t)⇤ = L⇥cos(t)⇤ = s
s2 + 1
.
Then
L

d
dt
cos(t)

= L

d
dt
f(t)

= sF (s) f(0)
= sF (s) cos(0) = sF (s) 1 = s s
s2 + 1
1 = 1
s2 + 1| {z }
L
⇥
sin(t)
⇤
ENG2011
Di↵erential Equations & Series
– 3 – lecture notes 2
Laplace Transforms II
(5) Verify rule 4 with f(t) = sinh(t).
First consider the left-hand side of 4
tZ
0
f(u) du =
tZ
0
sinh(u) du =
h
cosh(u)
it
0
= cosh(t) cosh(0)| {z }
1
= cosh(t) 1
Therefore
L
24 tZ
0
f(u) du
35 = Lhcosh(t) 1i = L⇥cosh(t)⇤| {z } L⇥1⇤|{z}
=
s
s2 1
1
s
=
s2 (s2 1)
(s2 1)s =
1
s (s2 1)
Now consider the right-hand side of 4
1
s
F (s) =
1
s
L⇥f(t)⇤ = 1
s
L⇥sinh(t)⇤ = 1
s
1
(s2 1) = left-hand side X
(6) Find the Laplace Transform of 2te3t.
We can do this using either 2 or using 5 .
Using the First Shift Theorem 2 , take f(t) = 2t so that
L⇥2te3⇤ = L⇥f(t)e3t⇤ = F (s+ 3)
From Table F (s) = L⇥2t⇤ = 2
s2
. Therefore
L⇥2te3⇤ = F (s+ 3) = 2
(s+ 3)2
Alternatively, using 5 , take g(t) = 2e3t so that
L⇥2te3t⇤ = L⇥tg(t)⇤ = d
ds
G(s)
Now, from Table,
G(s) = L⇥g(t)⇤ = L⇥2e3t⇤ = 2
s+ 3
Therefore
L⇥2te3t⇤ = d
ds
G(s) = d
ds
✓
2
s+ 3
◆
=
2
(s+ 3)2
ENG2011
Di↵erential Equations & Series
– 4 – lecture notes 2
Laplace Transforms II
(7) Find the Laplace Transform of the function
f(t) =
(
1 , t < 2
0 , t > 2
f
t0 2
f
1
First let us determine the Laplace Transform from first principles: we break down the integral over
the interval 0 6 t <1 into two parts; 0 6 t 6 2 and 2 < t <1.
L⇥ f(t) ⇤ = 1Z
0
f(t) est dt =
2Z
0
1 est dt +
1Z
2
0 est dt
| {z }
=0
=

1
s e
st
2
0
= 1
s
⇣
e2s 1
⌘
=
1
s
⇣
1 e2s
⌘
= F (s)
We can derive the same result using the Heaviside function and rule 1 :
We can write
f(t) = 1H(t2) =
(
1 , t < 2
0 , t > 2
Therefore
L⇥f(t)⇤ = Lh1H(t2)i
= L⇥1⇤ L⇥H(t2)⇤ = 1
s
1
s
e2s =
1
s
⇣
1 e2s
⌘
= F (s)
ENG2011
Di↵erential Equations & Series
– 5 – lecture notes 2
Laplace Transforms II
(8) Find the Laplace Transform of the function
f(t) =
(
t , 0 6 t 6 1
0 , t > 1
f
t0 1
1
From first principles; break down the integral 0 6 t <1 into two parts; 0 6 t 6 1 and 1 < t <1.
L⇥ f(t) ⇤ = 1Z
0
f(t) est dt =
1Z
0
t est dt +
1Z
1
0 est dt
| {z }
=0
Then evaluate the non-zero contribution (0 6 t 6 1) using integration by parts
L⇥ f(t) ⇤ = 1Z
0
t|{z}
u
est|{z}
v0
dt
=
h
t|{z}
u

1
s e
st| {z }
v
i1
0

1Z
0
1|{z}
u0

1
s e
st| {z }
v
dt
= 1s es
h
1
s2 e
st
i1
0
= 1s es

1
s2 e
s 1s2

= 1s2 es

1
s +
1
s2

= F (s)
We can derive the same result using the Heaviside function and rules 1 and 5 .
Noting that
1H(t1) =
(
1 , 0 6 t 6 1
0 , t > 1
we can write
f(t) = t
⇣
1H(t1)
⌘
Then
L⇥f(t)⇤ = Lht⇣1H(t1)⌘i
= L
h
t tH(t1)
i
= L⇥t⇤|{z} L⇥tH(t 1)⇤| {z }
=
1
s2

✓
d
ds
L⇥H(t 1)⇤◆
=
1
s2
+
d
ds
✓
1
s
es
◆
=
1
s2
es
✓
1
s
+
1
s2
◆
ENG2011
Di↵erential Equations & Series
– 6 – lecture notes 2
Laplace Transforms II
To finish this lecture we outline the proofs of the rules we have been using.
Proofs Of Rules 1 – 6
1 Linearity
L⇥af(t) + bg(t)⇤ = 1Z
0
⇣
af(t) + bg(t)
⌘
est dt
= a
1Z
0
f(t) est dt
| {z }
+ b
1Z
0
g(t) est dt
| {z }
= a L⇥f(t)⇤ + b L⇥g(t)⇤
2 First Shift Theorem
L⇥eatf(t)⇤ = 1Z
0
eatf(t) est dt =
1Z
0
f(t) e(sa)t dt
| {z }
= F (s a)
3 Derivatives
L⇥f 0(t)⇤ = 1Z
0
f 0(t) est dt =
h
f(t) est
i1
0| {z }
1Z
0
f(t)
s est dt (integration by parts)
= f(0) + s
1Z
0
f(t) est dt
| {z }
= f(0) + s F (s)
Similarly L⇥f 00(t)⇤ = s2F (s) sf(0) f 0(0) can be derived using integration by parts twice.
ENG2011
Di↵erential Equations & Series
– 7 – lecture notes 2
Laplace Transforms II
4 Integral
L
24 tZ
0
f(u) du
35 = 1Z
0
✓
tR
0
f(u) du
◆
| {z }
u
est|{z}
v0
dt
=
"✓
tR
0
f(u) du
◆
| {z }
u
✓
1
s e
st
◆
| {z }
v
#1
0| {z }

1Z
0
f(t)|{z}
u0
✓
1
s e
st
◆
| {z }
v
dt
= 0 +
1
s
1Z
0
f(t) est dt
| {z }
=
1
s
F (s)
5 Multiplication by t
L⇥tf(t)⇤ = 1Z
0
tf(t) est dt =
1Z
0
f(t)
⇣
t est
⌘
| {z } dt
=
1Z
0
f(t)
z }| {⇣
ddsest
⌘
dt
= d
ds
1Z
0
f(t) est dt = d
ds
F (s)
6 Second Shift Theorem
L⇥H(t a)f(t a)⇤ = 1Z
0
H(t a)f(t a) est dt =
1Z
a
f(t a) est dt
substitute
u = t a
=
1Z
0
f(u) es(u+a) du
= eas
1Z
0
f(u) esu du
| {z }
= eas F (s)
ENG2011
Di↵erential Equations & Series
– 8 – lecture notes 2
Laplace Transforms II
DIFFERENTIAL EQUATIONS & SERIES
LECTURE 3
THE INVERSE LAPLACE TRANSFORM
In lectures 1 & 2 we learnt about the Laplace Transform
f (t) F (s) = L⇥f (t)⇤ = 1Z
0
f (t) est dtL
Given a function f(t) we can compute the transform function F (s) – either from first principles, or
by using our Table of Transforms and the general results 1 – 6 presented in lecture 2.
When we use Laplace Transforms to solve ODEs (next lecture) what we need to do is ‘the opposite’.
That is, given a function F (s) we will need to find the corresponding function f(t).
This leads to the idea of the Inverse Laplace Transform.
L1⇥F (s)⇤ = f (t) F (s)L1
To determine the Inverse transform f(t) = L1⇥F (s)⇤ we again use of our Table of Transforms and
the general results 1 – 6 . For ease of reference these are summarised on the next page.
ENG2011
Di↵erential Equations & Series
– 1 – lecture 3
Inverse Laplace Transforms
Summary of Results on Laplace Transforms
L⇥f(t)⇤ = 1Z
0
f(t)estdt = F (s)
f(t) F (s)
1
1
s
t
1
s2
tn
n = 1, 2, 3, . . .
n!
sn+1
eat
1
s a
sin(at)
a
s2 + a2
cos(at)
s
s2 + a2
sinh(at)
a
s2 a2
cosh(at)
s
s2 a2
H(t a) 1
s
eas
Table of Transforms
1 Linearity
L⇥af(t) + b g(t)⇤ = aL⇥f(t)⇤+ bL⇥g(t)⇤
2 First Shift Theorem
L⇥eatf(t)⇤ = F (s a)
3 Derivatives
L⇥f 0(t)⇤ = sF (s) f(0)
L⇥f 00(t)⇤ = s2F (s) sf(0) f 0(0)
4 Integral
L
Z t
0
f(u) du

=
1
s
F (s)
5 Multiplication by t
L⇥tf(t)⇤ = d
ds
F (s)
6 Second Shift Theorem
L⇥H(t a)f(t a)⇤ = easF (s)
It is useful to write of the two Shift Theorems 2 , 6 in term of the inverse operator L1
2 L1
h
F (s a)
i
= eatf(t)
6 L1⇥easF (s)⇤ = H(t a)f(t a)
In the remaining part of this lecture we go through examples of finding inverse transforms.
These examples are divided into
• Inverse Transforms Of Simple Functions
• Inverse Transforms Using Partial Fractions
• Inverse Transforms Using First Shift Theorem And Completing The Square
ENG2011
Di↵erential Equations & Series
– 2 – lecture 3
Inverse Laplace Transforms
Inverse Transforms Of Simple Functions•
(1) Find the Inverse Laplace Transform of F (s) =
1
s2
.
We can see the answer directly from the table. Since L⇥ t ⇤ = 1
s2
it follows that t = L1

1
s2

(2) Find the Inverse Laplace Transform of F (s) =
3
s2
.
Here we use the table and the idea of linearity 1 . Write
L1

3
s2

= L1

3⇥ 1
s2

= 3L1

1
s2

= 3t
(3) Find the Inverse Laplace Transform of F (s) =
1
s3
.
From the table we know that L1

2
s3

= t2. So write
L1

1
s3

= L1

1
2 ⇥
2
s3

= 12 L
1

2
s3

= 12 t
2
(4) Find the Inverse Laplace Transform of F (s) =
1
s+ 4
.
Again we can read the answer directly from the table.
L1

1
s+ 4

= e4t (a = 4)
(5) Find the Inverse Laplace Transform of F (s) =
4
s 5.
Write
L1

4
s 5

= L1

4⇥ 1
s 5

= 4L1

1
s 5

= 4 e5t
(6) Find the Inverse Laplace Transform of F (s) =
3s 1
s2 + 4
.
Here we break down the function into two parts. Write
L1

3s 1
s2 + 4

= L1

3s
s2 + 4
1
s2 + 4

= L1

3s
s2 + 4

L1

1
s2 + 4

= 3L1

s
s2 + 4

12 L
1

2
s2 + 4

= 3 cos(2t) 12 sin(2t)
ENG2011
Di↵erential Equations & Series
– 3 – lecture 3
Inverse Laplace Transforms
Inverse Transforms Using Partial Fractions•
Sometimes we need to use partial fractions to rewrite the function F (s).
The rules for partial fractions (given on the Formula Sheet) are summarised by
p(x)
q(x)|{z}
deg(p)=
p(x)
(x+ ↵) (x+ )2(ax2 + bx+ c)
⌘ A
(x+ ↵)
+
Bx+ C
(x+ )2
+
Dx + E
(ax2 + bx+ c)
Here are some examples:
(7) Find the Inverse Laplace Transform of F (s) =
4s
s2 2s 3.
Write
4s
s2 2s 3 =
4s
(s+ 1)(s 3) =
A
s+ 1
+
B
s 3 = · · · =
1
s+ 1
+
3
s 3
so that
L1

4s
s2 2s 3

= L1

1
s+ 1
+
3
s 3

= L1

1
s+ 1

+ 3L1

1
s 3

= et + 3 e3t
(8) Find the Inverse Laplace Transform of F (s) =
s+ 1
s2 + 3s
.
Write
s+ 1
s2 + 3s
=
s+ 1
s(s+ 3)
=
A
s
+
B
s+ 3
= · · · = 1/3
s
+
2/3
s+ 3
so that
L1

s+ 1
s2 + 3s

= L1

1/3
s
+
2/3
s+ 3

= 13 L
1

1
s

+ 23 L
1

1
s+ 3

= 13 +
2
3 e
3t
(9) Find the Inverse Laplace Transform of F (s) =
1
(s 1)(s2 + 4).
Write
1
(s 1)(s2 + 4) =
A
s 1 +
Bs+ C
s2 + 4
= · · · = 1/5
s 1 +
1/5 s 1/5
s2 + 4
so that
L1

1
(s 1)(s2 + 4)

= 15 L
1

1
s 1
s+ 1
s4 + 4

= 15 L
1

1
s 1

| {z } 15 L
1

s
s2 + 4

| {z } 15 L
1

1
s2 + 4

| {z }
= 15 e
t 15 cos(2t) 15 ⇥ 12 sin(2t)
ENG2011
Di↵erential Equations & Series
– 4 – lecture 3
Inverse Laplace Transforms
Inverse Transforms Using First Shift Theorem and Completing the Square•
The First Shift Theorem 2 L⇥eatf(t)⇤ = F (s a) can be written as
L1⇥F (s a)⇤ = eatf(t) = eat L1⇥F (s)⇤
(10) Find the Inverse Laplace Transform of the function
2
(s 5)3 .
If we take F (s) =
2
s3
then
2
(s 5)3 = F (s 5) so that
L1

2
(s 5)3

= L1
h
F (s 5)
i
= e5tf(t)
Now
f(t) = L1
h
F (s)
i
= L1

2
s3

= t2 (from table)
Therefore
L1

2
(s 5)3

= e5t t2
(11) Find the Inverse Laplace Transform of F (s) =
s 2
s2 4s+ 5.
The denominator s2 4s+ 5 does not factorise so we cannot use partial fractions.
Instead we complete the square: s2 4s+ 5 = (s 2)2 + 1 so that
L1

s 2
s2 4s+ 5

= L1

s 2
(s 2)2 + 1

Now we use the First Shift Theorem: Replace s 2 with s (numerator and denominator) and
write
L1

s 2
(s 2)2 + 1

= e2t L1

s
s2 + 1

= e2t cos(t)
(12) Find the Inverse Laplace Transform of F (s) =
s+ 2
s2 + 6s+ 13
.
Complete the square: s2 + 6s+ 13 = (s+ 3)2 + 4 so that
L1

s+ 2
s2 + 6s+ 13

= L1

s+ 2
(s+ 3)2 + 4

To use the First Shift Theorem replace s+ 3 with s (numerator and the denominator)
Write
L1

s+ 2
(s+ 3)2 + 4

= L1

(s+ 3) 1
(s+ 3)2 + 4

= L1

(s+ 3)
(s+ 3)2 + 4

| {z } L
1

1
(s+ 3)2 + 4

| {z }
= e3t L1

s
s2 + 4

e3t L1

1
s2 + 4

= e3t
⇣
cos(2t) 12 sin(2t)
⌘
ENG2011
Di↵erential Equations & Series
– 5 – lecture 3
Inverse Laplace Transforms
DIFFERENTIAL EQUATIONS & SERIES
LECTURE 4
SOLVING ODEs USING LAPLACE TRANSFORMS I
In this lecture we learn how to use Laplace Transforms to solve linear ODEs.
The motivation for doing this was given in lecture 1.
We consider second-order, linear ODEs with constants coecients (a, b, c)
a x¨ + b x˙ + c x = (1)
There will be initial conditions for this equation: x(0) = x0, x˙(0) = v0, where x0, v0 are given values.
The Method Of Solution
. Take the Laplace Transform of equation (1)
L⇥a x¨ + b x˙ + c x⇤ = L []
. Use Linearity to write this as
aL⇥x¨⇤ + bL⇥x˙⇤ + cL⇥x⇤ = L [] (2)
. Define the Laplace Transform functions
(s) = L⇥(t)⇤ and X(s) = L⇥x(t)⇤
Note
• the function (t), the forcing term, will be known.
Therefore (s) can be computed; from first principles or the Table of Transforms etc.
• the function x(t), the solution to equation (1), is unknown.
The idea is to find X and then use Inverse Laplace Transforms to find x = L1⇥X⇤.
. Use rule 3 for derivatives to write
L [x˙] = sX x(0)
= sX x0
L⇥x¨⇤ = s2X s x(0) x˙(0)
= s2X s x0 v0
. Substitute these expressions into equation (2) and rearrange for X(s).
. Find x(t) using Inverse Laplace Transforms
x(t) = L1⇥X(s)⇤
ENG2011
Di↵erential Equations & Series
– 1 – lecture 4
ODEs I
Let’s do some examples:
In this lecture we consider examples with simple forcing functions (t).
In the next lecture we consider examples with more complicated forcing functions.
The following examples could be solved using the method taught in ENG1001.
You should check that we get the same answers.
(1) Find the solution of the homogeneous ODE
x¨ + 5 x˙ + 6 x = 0 , x(0) = 1 , x˙(0) = 0
Take Laplace Transforms to get
L⇥x¨⇤ + 5L⇥x˙⇤ + 6L⇥x⇤ = L⇥0⇤ = 0 , (⇤)
Use the rule for derivatives to get
L [x˙] = sX x(0)
= sX 1
L⇥x¨⇤ = s2X s x(0) x˙(0)
= s2X s 1 0
Substitute these into (⇤)
s2X s + 5 sX 1 + 6X = 0
or
s2 + 5s+ 6

X s 5 = 0
Rearrange for X
X =
s+ 5
s2 + 5s+ 6
Therefore
x(t) = L1⇥X(s)⇤ = L1 s+ 5
s2 + 5s+ 6

Now use the techniques developed in lecture 3 to evaluate this inverse transform.
s+ 5
s2 + 5s+ 6
=
s+ 5
(s+ 2)(s+ 3)
=
A
s+ 2
+
B
s+ 3
= · · · = 3
s+ 2
2
s+ 3
So
x(t) = L1

3
s+ 2
2
s+ 3

= 3L1

1
s+ 2

2L1

1
s+ 3

= 3e2t 2e3t
ENG2011
Di↵erential Equations & Series
– 2 – lecture 4
ODEs I
(2) Find the solution of the inhomogeneous ODE
x¨ 3 x˙ + 2 x = 2 et , x(0) = 2 , x˙(0) = 1
Take Laplace Transforms to get
L⇥x¨⇤ 3L⇥x˙⇤ + 2L⇥x⇤ = L⇥2 et⇤ = 2 1
s+ 1
, (⇤)
Use the rule for derivatives to get
L [x˙] = sX x(0)
= sX 2
L⇥x¨⇤ = s2X s x(0) x˙(0)
= s2X s 2 + 1
Substitute these into (⇤)
s2X 2s+ 1 3 sX 2 + 2X = 2
s+ 1
or
s2 3s+ 2X 2s+ 7 = 2
s+ 1
Rearrange for X

s2 3s+ 2X = 2
s+ 1
+ 2s 7 = 2 + (2s 7)(s+ 1)
s+ 1
=
2s2 5s 5
s+ 1
So
X =
2s2 5s 5
(s+ 1)(s2 3s+ 2)
Therefore
x(t) = L1⇥X(s)⇤ = L1 2s2 5s 5
(s+ 1)(s2 3s+ 2)

Now use the techniques developed in lecture 3 to evaluate this inverse transform.
2s2 5s 5
(s+ 1)(s2 3s+ 2) =
2s2 5s 5
(s+ 1)(s 2)(s 1) =
A
s+ 1
+
B
s 2 +
C
s 1
=
1/3
s+ 1
7/3
s 2 +
4
s 1
So
x(t) = 13 L
1

1
s+ 1

73 L
1

1
s+ 2

+ 4L1

1
s 1
= 13e
t 73e2t + 4et
ENG2011
Di↵erential Equations & Series
– 3 – lecture 4
ODEs I
(3) Find the solution of the resonant ODE
x¨ + 4 x˙ + 4x = te2t , x(0) = 1 , x˙(0) = 2
Take Laplace Transforms to get
L⇥x¨⇤ + 4L⇥x˙⇤ + 4L⇥x⇤ = L⇥t e2t⇤ (⇤)
Use general result L⇥tf(t)⇤ = ddsF (s) to evaluate the right-hand side
f(t) = e2t ) F (s) = 1
s+ 2
= (s+ 2)1
so that
L⇥t e2t⇤ = d
ds
(s+ 2)1 =
1
(s+ 2)2
Use the rule for derivatives to get
L [x˙] = sX x(0)
= sX 1
L⇥x¨⇤ = s2X s x(0) x˙(0)
= s2X s 1 2
Substitute these into (⇤)
s2X s 2 + 4 sX 1 + 4X = 1
(s+ 2)2
or
s2 + 4s+ 4
X s 6 = 1
(s+ 2)2
Rearrange for X
s2 + 4s+ 4
| {z }
(s+2)2
X =
1
(s+ 2)2
+ s+ 6
So
X =
1
(s+ 2)4
+
s+ 6
(s+ 2)2
Therefore
x(t) = L1⇥X(s)⇤ = L1 1
(s+ 2)4
+ L1

s+ 6
(s+ 2)2
Now use the techniques developed in lecture 3 to evaluate these inverse transforms
L1

1
(s+ 2)4
= e2tL1

1
s4
| {z }
first shift theorem
= e2t
1
3!
L1

3!
s4

= e2t
1
6
t3
and
L1

s+ 6
(s+ 2)2
= L1

(s+ 2) + 4
(s+ 2)2
= L1

1
s+ 2
+ L1

4
(s+ 2)2
= e2t + 4 e2t t
Combining
x(t) = e2t
⇣
1 + 4t+ 16 t
3
⌘
In the next lecture we will do further examples with more complicated forcing terms (t).
ENG2011
Di↵erential Equations & Series
– 4 – lecture 4
ODEs I
DIFFERENTIAL EQUATIONS & SERIES
LECTURE 5
SOLVING ODEs USING LAPLACE TRANSFORMS II
In lecture 4 we learnt how to use Laplace Transforms to solve ODEs.
In lecture 5 we develop this application further.
We consider 2 extensions:
• The application to ODEs with discontinuous and/or non-smooth forcing.
• The application to Systems of ODEs.
Solving ODEs with Discontinuous/Non-Smooth Forcing•
Again we illustrate the method using the second-order, linear ODE with constant coecients
a x¨ + b x˙ + c x = (1)
In many engineering systems the forcing term (t) can exhibit complicated behaviour. In particular
there may be discontinuous or non-smooth features, as illustrated in the following figure.
t t t
Figure 1
In such cases equation (1) can still be solve using Laplace Transforms.
The method presented in the last lecture still applies.
We will be making regular use of the Second Shift Theorem 6
L
h
f(ta)H(ta)
i
= eas F (s)
or, written in terms of the inverse operator L1,
f(ta)H(ta) = L1
h
eas F (s)
i
Some examples are presented on the following pages:
ENG2011
Di↵erential Equations & Series
– 1 – lecture 5
ODEs II
(1) Find the solution to the ODE
x¨ + 5 x˙ + 6 x = , x(0) = 1 , x˙(0) = 0
with
(t) =
(
1 , 0 6 t 6 1
0 , t > 1
Take Laplace Transforms to get
L⇥x¨⇤ + 5L⇥x˙⇤ + 6L⇥x⇤ = L⇥⇤ = , (⇤)
As before, use the rule for derivatives to get
L [x˙] = sX x(0)
= sX 1
L⇥x¨⇤ = s2X s x(0) x˙(0)
= s2X s 1 0
There are 2 ways to determine the transform of the forcing term:
From first principles
= L [] =
1Z
0
(t) est dt =
1Z
0
1 est dt =
1
s
1
s
es
Alternatively, write = 1H(t1) so that, from the Table of Transforms,
= L⇥1⇤ L⇥H(t1)⇤ = 1
s
1
s
es
Substitute these transforms into (⇤) to get
s2X s + 5 sX 1 + 6X = 1
s
1
s
es
or, rearranging for X,
X =
s+ 5
s2 + 5s+ 6
+
1
(s2 + 5s+ 6) s
1 es
Using partial fractions
s+ 5
s2 + 5s+ 6
=
s+ 5
(s+ 2)(s+ 3)
=
3
s+ 2
2
s+ 3
F1(s)
1
(s2 + 5s+ 6) s
=
1
(s+ 2)(s+ 3)s
=
1/2
s+ 2
+
1/3
s+ 3
+
1/6
s
F2(s)
So X(s) = F1(s) + F2(s)
1 es and
x(t) = L1⇥X(s)⇤ = L1⇥F1(s)⇤| {z }
f1(t)
+ L1⇥F2(s)⇤| {z }
f2(t)
L1⇥F2(s) es⇤| {z }
f2(t1)H(t1)
with
f1(t) = L1

3
s+ 2
L1

2
s+ 3
= 3 e2t 2 e3t
f2(t) = L1
 1/2
s+ 2
+ L1

1/3
s+ 3
+ L1

1/6
s
= 12e2t + 13e3t + 16
ENG2011
Di↵erential Equations & Series
– 2 – lecture 5
ODEs II
(2) Find the solution to the ODE
x¨ + 3 x˙ + 2 x = , x(0) = 0 , x˙(0) = 0
with
(t) =
(
t , 0 6 t 6 1
0 , t > 1
Take Laplace Transforms to get
L⇥x¨⇤ + 3L⇥x˙⇤ + 2L⇥x⇤ = L⇥⇤ = , (⇤)
Use the rule for derivatives to get
L [x˙] = sX x(0)
= sX
L⇥x¨⇤ = s2X s x(0) x˙(0)
= s2X
Write = t
1H(t1) so that, from the Table of Transforms and rule 5 (multiplication by t),
= L⇥t⇤ L⇥tH(t1)⇤ = 1
s2
+
d
ds
✓
1
s
es
◆
=
1
s2
✓
1
s
+
1
s2
◆
es =
1
s2
s+ 1
s2
es
Substitute these transforms into (⇤) to get
s2 + 3s+ 2
| {z }
(s+ 1)(s+ 2)
X =
1
s2
s+ 1
s2
es
Rearranging for X,
X =
1
s2(s+ 1)(s+ 2)
1
s2(s+ 2)
es
Using partial fractions
1
s2(s+ 1)(s+ 2)
= 1
4
✓
3s 2
s2
◆
+
1
s+ 1
1
4
✓
1
s+ 2
◆
F1(s)
1
s2(s+ 2)
=
1
4
✓
s+ 2
s2
◆
1
4
✓
1
s+ 2
◆
F2(s)
So X(s) = F1(s) F2(s) es and
x(t) = L1⇥X(s)⇤ = L1⇥F1(s)⇤| {z }
f1(t)
L1⇥F2(s) es⇤| {z }
f2(t1)H(t1)
(†)
with
f1(t) = 14 L
1

3s 2
s2
+ L1

1
s+ 1
14 L
1

1
s+ 2
= 14
3 2t + et 14e2t
f2(t) =
1
4 L
1

s+ 2
s2
14 L
1

1
s+ 2
= 14
1 + 2t
14e2t
Finally, substituting these into (†) gives
x(t) = f1(t) f2(t1)H(t1) =
n
14
32t+et 14e2to 14n⇣1+2(t1)⌘e2(t1)oH(t1)
ENG2011
Di↵erential Equations & Series
– 3 – lecture 5
ODEs II
Solving Systems of ODEs•
So far we have considered the application of Laplace Transforms to solving a single ODE.
Many engineering systems have to be modelled using more than one ODE, and these di↵erential
equations are coupled together. Here are two simple examples.
Figure 2 shows a model describing the oscillations of two coupled masses
k1 k2
↵1 ↵2
M1 M2
x1(t) x2(t)
(t)
Figure 2
The corresponding ODEs for the mass displacements x1(t), x2(t) are
M1 x¨1 = k1x1 ↵1x˙1 + k2(x2 x1) + ↵2(x˙2 x˙1)
M2 x¨2 = k2(x2 x1) ↵2(x˙2 x˙1) +
Figure 3 shows a mixing-tank flow system designed to mix a dissolved chemical.
The chemical, at a concentration x0 (kg/m3), enters T1 with a flow rate Q (m3/s).
Q
q2
q1
T1 T2
Q
Figure 3
The concentrations x1, x2 in the tanks vary due to the fluid exchanges (q1 q2 = Q), and are
modelled by the ODEs
V1 x˙1 = q1x1 + q2x2 + Qx0
V2 x˙2 = q1x1 (q2 +Q)x2
where V1, V2 (m3) are the volumes of the two tanks.
ENG2011
Di↵erential Equations & Series
– 4 – lecture 5
ODEs II
We can use Laplace Transforms to solve coupled, linear ODEs.
Here are some examples.
(1) Find the functions x1(t), x2(t) satisfying the coupled first-order di↵erential equations
x˙1 + x2 = 0 , x1(0) = 1 ,
x˙2 x1 = 0 , x2(0) = 0 .
Define X1 = L [x1] and X2 = L [x2]. Then, taking Laplace Transforms of both equations,
sX1 1 + X2 = 0 (1)
sX2 X1 = 0 (2)
From (1), X2 = sX1 + 1. Substitute this into (2) to get
s
sX1 + 1X1 = s2 + 1X1 + s = 0
so that
X1 =
s
s2 + 1
and then, from (2),
X2 =
1
s
X1 =
1
s2 + 1
Therefore
x1 = L1
⇥
X1
⇤
= L1

s
s2 + 1

= cos(t)
and
x2 = L1
⇥
X2
⇤
= L1

1
s2 + 1
= sin(t)
(2) Solve the the coupled, di↵erential equations
x˙1 + 2x1 x2 = 0 , x1(0) = 0 ,
x˙2 8x1 = 0 , x2(0) = 1 .
Take Laplace Transforms and rearange to get
X1 =
1
s2 + 2s 8 , X2 =
s+ 2
s2 + 2s 8 .
Complete The Square to evaluate the Inverse Transforms;
x1 = L1 [X1] = L1

1
(s+ 1)2 9
= 13e
t sinh(3t)
x2 = L1 [X2] = L1

(s+ 1) + 1
(s+ 1)2 9
= et cosh(3t) + 13 sinh(3t)
Alternatively, write s2 + 2s 8 = (s 2)(s+ 4) and use partial fractions.
[Use cosh(x) = 12 (e
x + ex), sinh(x) = 12 (e
x ex), to obtain the form of answer above.]
ENG2011
Di↵erential Equations & Series
– 5 – lecture 5
ODEs II
(3) Solve the coupled, di↵erential equations
x˙1 + x2 = 1 =
(
1 , 0 6 t 6 3
0 , t > 3
x1(0) = 0
x˙2 + 4x1 = 2 =
(
0 , 0 6 t 6 3
1 , t > 3
x2(0) = 0
The right-hand side functions 1, 2 can be written using Heaviside functions
1 = 1H(t3) , 2 = H(t3) ,
so that
1 = L [1] = 1
s
1
s
e3s , 2 = L [2] = 1
s
e3s .
The Transforms of the di↵erential equations are then
sX1 + X2 =
1
s
1
s
e3s (1)
sX2 + 4X1 =
1
s
e3s (2)
Solving for X1 we get (after a bit of algebra)
X1 =
1
(s2 4)| {z }
F1(s)
1
(s2 4)| {z }
F1(s)
e3s 1
s(s2 4)| {z }
F2(s)
e3s
Therefore, using the Second Shift Theorem 6 ,
x1(t) = L1 [X1] = f1(t) f1(t3)H(t3) f2(t3)H(t3)
where
f1(t) = L1
h
F1(s)
i
= L1

1
(s2 4)
= 12 sinh(2t) =
1
4
e2t e2t
f2(t) = L1
h
F2(s)
i
= L1

1
s(s2 4)
= L1

1/4
s
+
1/8
s 2 +
1/8
s+ 2
partial
fractions
= 14 + 18 e2t + 18e2t = 14 + 14 cosh(2t)
To determine x2(t) rearrange (1)
X2 = sX1 + 1 = s
(s2 4)| {z }
F3(s)
+
s
(s2 4)| {z }
F3(s)
e3s +
1
(s2 4)| {z }
F1(s)
e3s + 1
so that
x2(t) = L1 [X2] = f3(t) + f3(t3)H(t3) + f1(t3)H(t3) + 1
where
f3(t) = L1
h
F3(s)
i
= L1

s
(s2 4)
= cosh(2t)
ENG2011
Di↵erential Equations & Series
– 6 – lecture 5
ODEs II
DIFFERENTIAL EQUATIONS & SERIES
LECTURE 6
TAYLOR SERIES
In Stage 1 Engineering Maths (ENG1001) we learnt about Maclaurin Series :
The idea is to represent a function f(x) as series of powers of x.
This leads to the general formula
f(x) = f(0) + f 0(0) x +
1
2!
f 00(0) x2 + · · · =
1X
n=0
1
n!
f (n)(0) xn (⇤)
and the Maclaurin Series for standard functions
series converges for
ex = 1 + x +
1
2!
x2 +
1
3!
x3 +
1
4!
x4 + · · · all x
sin(x) = x 1
3!
x3 +
1
5!
x5 1
7!
x7 + · · · all x (in radians)
cos(x) = 1 1
2!
x2 +
1
4!
x4 1
6!
x6 + · · · all x (in radians)
ln(1 + x) = x 1
2
x2 +
1
3
x3 1
4
x4 + · · · 1 < x 6 +1
(1 + x)b = 1 + bx + b(b1)2! x
2 + b(b1)(b2)3! x
3 + · · · 1 < x < +1
(unless b is a positive integer)
The Taylor Series of a function f(x) is a generalisation of this idea.
Instead of (⇤) we find an expansion in a series of powers of (x a), where a is some constant.
This leads to
f(x) = f(a) + f 0(a) (x a) + 1
2!
f 00(a) (x a)2 + · · · =
1X
n=0
1
n!
f (n)(a) (x a)n
We call this the Taylor Series expansion of f(x) about x = a.
Notice that the Maclaurin Series formula (⇤) is a special case of this, with a = 0.
ENG2011
Di↵erential Equations & Series
– 1 – lecture 6
Taylor Series
讲 a 侧了
We can use the Maclaurin Series for standard functions to construct corresponding Taylor Series.
Here are some examples:
(1) Find the Taylor Series of ex about x = 2.
We can write ex = e2+(x2) = e2 ⇤ e(x2).
Now use the standard Maclaurin Series for ex, replacing x with (x2).
ex = e2+(x2) = e2 ⇤ e(x2) = e2
✓
1 + (x2) + 12! (x2)2 + 13! (x2)3 + · · ·
◆
(2) Find the Taylor Series of cos(x) about x = ⇡/3.
Write
cos(x) = cos
⇡/3+ (x⇡/3) = cos(⇡/3) cosx⇡/3 sin(⇡/3) sinx⇡/3
= 12 cos
x⇡/3 p32 sinx⇡/3
Now use the standard Maclaurin Series for cos(x) and sin(x), replacing x with (x⇡/3).
cos(x⇡/3) = 1 12 (x⇡/3)2 + 14! (x⇡/3)4 + · · ·
sin(x⇡/3) = (x⇡/3) + 13! (x⇡/3)3 + 15! (x⇡/3)5 · · ·
Combining
cos(x) = 12
⇣
1 12 (x⇡/3)2 + · · ·
⌘
p
3
2
⇣
(x⇡/3) + 13! (x⇡/3)3 + · · ·
⌘
= 12
p
3
2 (x⇡/3) 14 (x⇡/3)2
p
3
12 (x⇡/3)3 + · · ·
(3) Find the Taylor Series of ln(x+ 2) about x = 1.
Write
ln(x+ 2) = ln
1 + (x1) + 2 = ln3 + (x1) = ln✓3h1 + 13 (x1)i◆
= ln (3) + ln
⇣
1 + 13 (x1)
⌘
Now use the standard Maclaurin Series for ln(1 + x), replacing x with 13 (x1) to get
ln(x+ 2) = ln(3) + 13 (x1) 118 (x1)2 + 181 (x1)3 1324 (x1)4 + · · ·
The standard Maclaurin Series for ln(1 + x) converges for 1 < x 6 +1.
If follows that the Taylor Series we have derived converges for
1 < 13 (x1) 6 +1 that is 3 < x 6 4
ENG2011
Di↵erential Equations & Series
– 2 – lecture 6
Taylor Series
L’Hoˆpital’s Rule
If functions f(x) and g(x) are both equal to 0 at the same point x = a, that is f(a) = g(a) = 0, then
what is the value of the limit
lim
x!a
f(x)
g(x)
?
We cannot simply put x = a since
f(a)
g(a)
=
0
0
is undetermined.
L’Hoˆpital’s Rule states that
lim
x!a
f(x)
g(x)
= lim
x!a
f 0(x)
g0(x)
if
f
g
! 0
0
= lim
x!a
f 00(x)
g00(x)
if
f 0
g0
! 0
0
We can prove this rule using Taylor Series expansions: Expand f(x) about x = a,
f(x) = f(a)|{z}
0
+ f 0(a) (x a) + 1
2!
f 00(a) (x a)2 + · · ·
= (x a)
⇣
f 0(a) +
1
2!
f 00(a) (x a) + · · ·
⌘
similarly
g(x) = (x a)
⇣
g0(a) +
1
2!
g00(a) (x a) + · · ·
⌘
Therefore, dividing and cancelling the common factor of (x a),
f(x)
g(x)
=
f 0(a) + 12! f
00(a) (x a) + · · ·
g0(a) + 12! g
00(a) (x a) + · · · !
f 0(a)
g0(a)
(as x! a)
If f 0(a) = g0(a) = 0 then we can repeat this process and cancel another common factor of (x a).
Let’s do some examples of using L’Hoˆpital’s Rule.
ENG2011
Di↵erential Equations & Series
– 3 – lecture 6
Taylor Series
(1) Find the value of
1 cos(x)
sin(x)
as x! 0.
f(x) = 1 cos(x) ! 0
as x! 0
g(x) = sin(x) ! 0
f 0(x) = sin(x) ! 0
as x! 0
g0(x) = cos(x) ! 1
Therefore, as x! 0,
f(x)
g(x)
=
1 cos(x)
sin(x)
! f
0(0)
g0(0)
=
0
1
= 0
(2) Find the value of
p
1 + xp1 x
3x
as x! 0.
f(x) =
p
1 + xp1 x ! 0
as x! 0
g(x) = 3x ! 0
f 0(x) =
1
2
p
1 + x
+
1
2
p
1 x ! 1
as x! 0
g0(x) = 3 ! 3
Therefore, as x! 0,
f(x)
g(x)
=
p
1 + xp1 x
3x
! f
0(0)
g0(0)
=
1
3
(3) Find the value of
1 ex2
x2
as x! 0.
f(x) = 1 ex2 ! 0
as x! 0
g(x) = x2 ! 0
f 0(x) = 2xex
2 ! 0
as x! 0
g0(x) = 2x ! 0
f 00(x) = (2 4x2)ex2 ! 2
as x! 0
g00(x) = 2 ! 2
Therefore, as x! 0,
f(x)
g(x)
=
1 ex2
x2
! f
00(0)
g00(0)
=
2
2
= 1
ENG2011
Di↵erential Equations & Series
– 4 – lecture 6
Taylor Series
DIFFERENTIAL EQUATIONS & SERIES
LECTURE 7
FOURIER SERIES
Motivation
Fourier Series play a central role in the mathematical analysis of waves and vibrations.
They are also the foundation of modern signal processing and our digital technology.
You would have no smart phone without Fourier Series analysis (and complex numbers).
Fourier Series were introduced in Stage 1 Engineering Maths (ENG1001).
In this lecture we revise Fourier Series.
The application of Fourier Series to Partial Di↵erential Equations is considered in lectures 8 & 9.
The basic idea is that of Harmonic Analysis :
We seek to decompose a given signal (function) f(t) into component sine and cosine functions.
k
+ + +
The component functions will have di↵erent, specified frequencies.
For each frequency !n we need to determine the corresponding amplitude An.
The amplitude tells us the contribution of the frequency to the original signal.
We can plot amplitudes against frequencies to show the frequency spectrum of the signal.
f (t)
t
An
!n
!
Signal Frequency Spectrum
The ability to identify dominant frequencies is an important design tool in many engineered systems.
ENG2011
Di↵erential Equations & Series
– 1 – lecture 7
Fourier Series
Mathematically we try to express the signal f(t) in the form of a Fourier Series
1X
n=0
an cos(!nt) + bn sin(!nt) (1)
Note that, using the trigonometric identity,
a cos(✓) + b sin(✓) ⌘ A cos(✓) amplitude A =
p
a2 + b2
phase shift = tan1
b
a
we can write series (1) in the form
1X
n=0
An cos
⇣
(!n(t ⌧n)
⌘
which shows how the Fourier Series coecients an, bn determine the amplitudes An =
p
a2n + b
2
n.
To simplify the mathematics we start by taking the frequencies !n = n = 0, 1, 2, 3, . . . and write
the corresponding form of equation (1) as
1
2a0 +
1X
n=1
an cos(nt) + bn sin(nt) (⇤)
Note that !0 = 0, and so the first term in series (1) becomes a constant.
This constant term has been written separately in (⇤) and the summation starts with n = 1.
The factor of 12 is included with a0 to simplify formulæ that are presented later.
Next note that
cos
n[t+ 2⇡]
= cos
nt+ 2⇡n
= cos
nt
sin
n[t+ 2⇡]
= sin
nt+ 2⇡n
= sin
nt
This means that the function given by (⇤) is periodic, with period 2⇡.
So we can use (⇤) to represent a function f(t) over an interval of length 2⇡. We choose ⇡ 6 t 6 +⇡.
The series will then represent f(t), ⇡ 6 t 6 +⇡, and it’s periodic extension outside of this interval.
ENG2011
Di↵erential Equations & Series
– 2 – lecture 7
Fourier Series
Here are examples of functions f(t), ⇡ 6 t 6 +⇡, and their periodic extensions over 3⇡ 6 t 6 +3⇡:
f(t) = t2
f
t
3⇡ 2⇡ ⇡ 0 ⇡ 2⇡ 3⇡
f(t) =
8<: 0 , ⇡ 6 t < 01 , 0 6 t < ⇡
f
t
03⇡ 2⇡ ⇡ ⇡ 2⇡ 3⇡
1
f(t) =
8>>><>>>:
1 +
1
⇡
t , ⇡ 6 t 6 0
1 1
⇡
t , 0 6 t 6 ⇡
f
t
03⇡ 2⇡ ⇡ ⇡ 2⇡ 3⇡
1
ENG2011
Di↵erential Equations & Series
– 3 – lecture 7
Fourier Series
Finding Fourier Coecients an, bn
Given a function f(t), ⇡ 6 t 6 +⇡, we need to determine the coecients an, bn such that
f(t) ⌘ 12a0 +
1X
n=1
an cos(nt) + bn sin(nt)
These coecients are given by the following formulæ
(the derivation of these formulæ is given at the end of this lecture)
an =
1
⇡
+⇡Z
⇡
f(t) cos(nt) dt bn =
1
⇡
+⇡Z
⇡
f(t) sin(nt) dt
n = 0, 1, 2, . . . n = 1, 2, . . .
Evaluating these integrals often requires integration by partsZ
u0v dt = uv
Z
uv0 dt
It is also useful to remember the following results
sin

±n⇡

= 0 cos

±n⇡

=
1n
f(t) odd f(t) even
f(t) = f(t) f(t) = +f(t)
an = 0 bn = 0
bn =
2
⇡
⇡Z
0
f(t) sin(nt) dt an =
2
⇡
⇡Z
0
f(t) cos(nt) dt
f(t)
f(t) t
t
f(t)f(t)
tt
ENG2011
Di↵erential Equations & Series
– 4 – lecture 7
Fourier Series
Let’s do some examples:
(1) Find the Fourier Series for the function f(t) = t, ⇡ < t < +⇡.
The graph of f(t), and its periodic extension to [3⇡,+3⇡], is
f
t
3⇡ 2⇡ ⇡
⇡ 2⇡ 3⇡
The function f(t) is odd; f(t) = f(t).
Therefore the Fourier coecients an are zero and f(t) =
1P
n=1
bn sin(nt).
bn =
1
⇡
+⇡Z
⇡
f(t) sin(nt)| {z }
even
dt =
2
⇡
⇡Z
0
f(t) sin(nt) dt =
2
⇡
⇡Z
0
t sin(nt) dt
=
2
⇡
8<:

1
n
t cos(nt)
⇡
0
+
1
n
⇡Z
0
cos(nt) dt
9=; = 2⇡
(
1
n
⇡ cos(n⇡)| {z }
(1)n
+
1
n

1
n
sin(nt)
⇡
0| {z }
=0
)
= 2 (1)
n
n
f(t) = 2
1X
n=1
(1)n+1
n
sin(nt) = 2
⇣
sin(t) 12 sin(2t) + 13 sin(3t) · · ·
⌘
(⇤)
The following figure shows the graph of f(t), and its periodic extension to [3⇡,+3⇡], together with
the approximation given by the first 8 terms in the Fourier Series representation (⇤).
0 50 100 150 200 250 300 350 400
0
50
100
150
200
3⇡ 2⇡ ⇡ ⇡ 2⇡ 3⇡
Notice that the function is discontinuous at t = ±⇡, ±3⇡, . . . and that, near these points, the Fourier
Series approximation is less accurate; exhibiting larger oscillations at these points of discontinuity.
This is a typical feature, know as Gibbs phenomenom.
Note also that at these points of discontinuity the value of the Fourier Series is 0. This is the average
of the left and right limits of the function at these points.
In general, at a point of discontinuity a the Fourier Series will converge to
1
2

lim
t! a
left
f(t) + lim
t! a
right
f(t)
!
ENG2011
Di↵erential Equations & Series
– 5 – lecture 7
Fourier Series
(2) Find the Fourier Series for the function f(t) = t2, ⇡ 6 t 6 +⇡.
The graph of f(t), and its periodic extension to [3⇡,+3⇡], is
f
t
3⇡ 2⇡ ⇡ 0 ⇡ 2⇡ 3⇡
The function f(t) is even; f(t) = f(t).
Therefore the Fourier coecients bn are zero, and f(t) =
1
2a0 +
1P
n=1
an cos(nt).
an =
1
⇡
+⇡Z
⇡
f(t) cos(nt)| {z }
even
dt =
2
⇡
⇡Z
0
f(t) cos(nt) dt =
2
⇡
⇡Z
0
t2 cos(nt) dt
a0 =
2
⇡
⇡Z
0
t2 dt =
2
⇡

1
3
t3
⇡
0
= 23 ⇡
2 , an =
2
⇡
(
1
n
t2 sin(nt)
⇡
0| {z }
=0
1
n
⇡Z
0
2t sin(nt) dt
)
= 4
⇡n
(
1
n
t cos(nt)
⇡
0
+
1
n
⇡Z
0
1 cos(nt) dt
| {z }
=0
)
=
4
n2
(1)n
Therefore
f(t) = 13⇡
2 + 4
1X
n=1
(1)n
n2
cos(nt) = 13⇡
2 + 4
⇣
cos(t) + 1
22
cos(2t) 1
32
cos(3t) + · · ·
⌘
If we set t = 0 then, since that f(0) = 0 and cos(0) = 1, this Fourier Series gives us the result
1
12⇡
2 = 1 1
22
+
1
32
1
42
· · ·
ENG2011
Di↵erential Equations & Series
– 6 – lecture 7
Fourier Series
(3) Find the Fourier Series of the function
f(t) =
8<: 0 , ⇡ < t < 01 , 0 6 t < ⇡
The graph of f(t), and its periodic extension to [3⇡,+3⇡], is
f
t
03⇡ 2⇡ ⇡ ⇡ 2⇡ 3⇡
1
The function is neither odd or even.
a0 =
1
⇡
+⇡Z
⇡
f(t) dt =
1
⇡
⇡Z
0
1 dt = 1
an =
1
⇡
+⇡Z
⇡
f(t) cos(nt) dt =
1
⇡
⇡Z
0
1 cos(nt) dt =
1
⇡

1
n
sin(nt)
⇡
0
= 0
bn =
1
⇡
+⇡Z
⇡
f(t) sin(nt) dt
=
1
⇡
⇡Z
0
1 sin(nt) dt =
1
⇡

1
n
cos(nt)
⇡
0
= 1
⇡n
⇣
(1)n 1
⌘
=
8><>:
0 n even
2
⇡n
n odd
The formula for non-zero bn (n odd) can be written b2k1 =
2
⇡(2k 1) , k = 1, 2, 3, . . ..
Therefore
f(t) =
1
2
+
2
⇡
1X
k=1
1
2k 1 sin
(2k 1)t
Note that an = 0 (all n 6= 0) and so there are no cos(nt) contributions to this series.
To explain this consider the shifted function
g(t) = f(t) 1
2
The function g(t) is odd, and it’s Fourier coecients, bn, are the same as those for f(t).
If we spot that a function f(t) can be made odd or even by a suitable shifting of axes then we can
avoid unnecessary integrations.
Again note that the value of the Fourier Series at the point of discontinuity (t = 0) is 1/2, which is
the average value of the left and right limits of f(t) at this point.
ENG2011
Di↵erential Equations & Series
– 7 – lecture 7
Fourier Series
A Couple Of Generalizations
• So far we have only considered functions f(t) defined on the interval ⇡ < t < +⇡.
If f(t) is defined on the interval ↵ < t < +↵ then we obtain the Fourier Series by introducing
the scaling
⌧ =
⇡
↵
t
so that ⌧ = ±⇡ when t = ±↵.
Write f(t) = f

T
↵ ⌧

= g(⌧).
The Fourier Coecients for f(t) are then obtained from those for g(⌧). This gives
an =
1
↵
+↵Z
↵
f(t) cos
⇣n⇡
↵
t
⌘
dt bn =
1
↵
+↵Z
↵
f(t) sin
⇣n⇡
↵
t
⌘
dt
• The Fourier series of a function F (t) with period T can be written
1
2a0 +
1X
n=1
an cos(n!t) + bn sin(n!t) ,
an =
2
T
t0+TR
t0
F (t) cos(n!t) dt ,
bn =
2
T
t0+TR
t0
F (t) sin(n!t) dt ,
where ! = 2⇡T , and t0 is any convenient lower limit of integration.
ENG2011
Di↵erential Equations & Series
– 8 – lecture 7
Fourier Series
Half-Range Sine And Cosine Fourier Series
We sometimes want to represent a function f(t) as a series that involves only sin( ) or cos( ) functions.
These are called Fourier Sine and Fourier Cosine Series.
When can do this by considering an extended function ef(t), which is designed to be either odd or
even. Here are two examples.
(1) Find the Fourier Sine Series of the function f(t) = 1, 0 < t < ⇡.
Define the extended function ef(t), ⇡ < t < +⇡, by
ef(t) =
8<: 1 , ⇡ < t < 01 , 0 < t < +⇡
The function ef is the same as f on the interval (0,⇡), and so the Fourier Series for ef provides a
representation for f . Further, this Fourier Series involves only sin( ) functions since ef is odd and
therefore an = 0 (all n).
bn =
1
⇡
+⇡Z
⇡
ef(t) sin(nt)| {z }
even
dt
=
2
⇡
⇡Z
0
1 sin(nt) dt =
2
⇡

1
n
cos(nt)
⇡
0
= 2
⇡n
⇣
(1)n 1
⌘
=
8><>:
0 n even
4
⇡n
n odd
Therefore
f(t) = 1 =
4
⇡
1X
k=1
1
2k 1 sin
(2k 1)t (0 < t < ⇡)
(2) Find the Fourier Cosine Series of the function f(t) = t, 0 < t < ⇡.
Define the extended function ef(t), ⇡ < t < +⇡, by
ef(t) =
8<: t , ⇡ < t < 0t , 0 < t < +⇡
⇡ +⇡
⇡
t
The function ef is even and therefore bn = 0.
an =
1
⇡
+⇡Z
⇡
ef(t) cos(nt)| {z }
even
dt =
2
⇡
⇡Z
0
t cos(nt) dt =
8>>>><>>>>:
⇡ n = 0
0 n 6= 0 even
4
⇡n2
n 6= 0 odd
Therefore
f(t) = t = 12⇡
4
⇡
1X
k=1
1
(2k 1)2 cos

(2k 1)t (0 < t < ⇡)
ENG2011
Di↵erential Equations & Series
– 9 – lecture 7
Fourier Series
Derivation Of Formulæ For an, bn
To derive the formulæ for the Fourier coecients an and bn we use the following important results
For all m,n = 1, 2, 3, . . .
+⇡Z
⇡
sin(mt) cos(nt) dt = 0 , and
+⇡Z
⇡
cos(mt) cos(nt) dt
+⇡Z
⇡
sin(mt) sin(nt) dt
9>>>>>>>=>>>>>>>;
=
8<: 0 m 6= n ,⇡ m = n .
[Exercise 7.5 asks you to prove these results].
Write
f(t) = 12a0 +
1X
m=1
am cos(mt) + bm sin(mt) (⇤)
Multiply (⇤) by cos(nt) and integrate over [⇡,+⇡]
+⇡Z
⇡
f(t) cos(nt) dt = 12a0
+⇡Z
⇡
cos(nt) dt +
1X
m=1
am
+⇡Z
⇡
cos(mt) cos(nt) dt + bm
+⇡Z
⇡
sin(mt) cos(nt) dt
| {z }
0
= 12a0
+⇡Z
⇡
cos(nt) dt +
1X
m=1
am
+⇡Z
⇡
cos(mt) cos(nt) dt
If n = 0 this gives
+⇡Z
⇡
f(t) dt = 12a0
+⇡Z
⇡
1 dt = ⇡a0
while, with n = 1, 2, 3 . . . , we get
+⇡Z
⇡
f(t) cos(nt) dt = 12a0
+⇡Z
⇡
cos(nt) dt
| {z }
0
+
1X
m=1
am
+⇡Z
⇡
cos(mt) cos(nt) dt
| {z }z }| {
0 ⇡
m 6= n m = n
= ⇡an
Therefore, in all cases,
an =
1
⇡
+⇡Z
⇡
f(t) cos(nt) dt
Similarly, multiplying (⇤) by sin(nt) and integrating over [⇡,+⇡] gives
bn =
1
⇡
+⇡Z
⇡
f(t) sin(nt) dt
ENG2011
Di↵erential Equations & Series
– 10 – lecture 7
Fourier Series
DIFFERENTIAL EQUATIONS & SERIES
LECTURE 8
PARTIAL DIFFERENTIAL EQUATIONS I
Introduction
Many mathematical models take the form of equations for functions that depend on more than one
independent variable. These independent variables often define positions in space (x, y, z) and time t.
For example
The temperature ✓(x, y, z, t) in a solid.
The velocity u(x, y, z, t) in a fluid flow.
The displacement u(x, y, t) of a vibrating surface.
The equations for these functions usually involve rates of change, and therefore partial derivatives.
These equations are then referred to as Partial Di↵erential Equations (PDEs).
In the next two lectures we learn a method for solving linear PDEs. Here are some examples:
The Heat Conduction Equation
The rate of change of the temperature ✓(x, y, z, t) in a material due to conductive heat transport can
be modelled, using conservation of energy, by the PDE
@✓
@t
= r2✓ = 
✓
@2✓
@x2
+
@2✓
@y2
+
@2✓
@z2
◆
where  is the thermal di↵usion coecient. This is an example of a parabolic PDE.
The Linearized Wave Equation
Small vertical displacements u(x, y, t) of points in a thin vibrating membrane can be modelled, using
Newton’s Second Law, by
@2u
@t2
= a2r2u = a2
✓
@2u
@x2
+
@2u
@y2
◆
where a denotes a wave speed of propagating waves. This is an example of a hyperbolic PDE.
Laplace’s Equation
The PDE (Laplace’s Equation)
r2 = @
2
@x2
+
@2
@y2
+
@2
@z2
= 0
appears in many mathematical models. This is an example of an elliptic PDE.
The function (x, y, z) might represent a steady-state temperature distribution (consider the heat
conduction equation above as t ! 1 and @✓/@t ! 0), or a potential function defining the the fluid
velocity u = r in an irrotational (r⇥ u = 0) and incompressible (r ·u = 0) flow.
ENG2011
Di↵erential Equations & Series
– 1 – lecture 8
PDEs I
Method Of Solution: Separation Of Variables
Solutions to these types of PDEs can be constructed using a Separation Of Variables technique.
The basic idea is as follows:
Given a PDE for a function f(x, t) we look for solutions in the form of a product
f(x, t) = X(x)T (t) .
Substitute this form for f(x, t) into the PDE to obtain ODEs for the functions X(x) and T (t).
Solve these ODEs for X(x) and T (t) (Stage 1 Maths).
The general solutions X(x) and T (t) will involve arbitrary constants.
These constants can be determined from boundary and/or initial conditions associated with the PDE.
To illustrate this method we apply it, in this lecture, to 2 PDEs
• the one-dimensional wave equation,
• the one-dimensional heat equation.
The One-Dimensional Wave Equation
Small vertical displacements u(x, t) in a tensioned wire, fixed at end-points x = 0, x = ` can be
modelled by the one-dimensional linearized wave equation
@2u
@t2
= a2
@2u
@x2
, 0 6 x 6 ` , t > 0 , (1)
with constant wave speed a determined by a2 = ⌧/⇢, ⌧ tension, ⇢ wire density (mass/length).
boundary conditions
Corresponding to fixed end-points, the boundary conditions for this PDE specify zero displacements
at x = 0 and x = `,
u(0, t) = 0 , u(`, t) = 0 , t > 0 .
initial conditions
We can start the wire vibrating in various ways, which will determine the initial conditions.
We must specify the initial displacement and the initial velocity of each point along the wire,
u(x, 0) = 0(x) ,
0 6 x 6 ` ,
@u
@t
(x, 0) = 1(x) ,
where 0 and 1 are given functions.
Now look for solutions to (1) in a Separated Variable form
u(x, t) = X(x)T (t) .
Then
@2u
@t2
= X(x) T¨ (t) and
@2u
@x2
= X 00(x)T (t) .
ENG2011
Di↵erential Equations & Series
– 2 – lecture 8
PDEs I
Substitute these into (1) to get
X T¨ = a2X 00 T ,
and rearrange (separate variables)
1
a2
T¨
T
=
X 00
X
.
The left-hand side is a function of t only, while the right-hand side is a function of x only.
It follows that both sides must be constant (not dependent on x or t). [This is the key point.]
Later, by considering the physics, we will see that this constant must be negative.
Therefore, we choose to denote the constant by 2. (We will need to determine the value of .)
So now
1
a2
T¨
T
=
X 00
X
= 2 .
This defines two ODEs, for X and T :
X 00 + 2X = 0 , (2)
T¨ + a22T = 0 . (3)
We can solve these second-order linear ODEs using Stage 1 Maths:
Equation (2)
To solve (2) look for solutions in the form X(x) = emx, giving the auxiliary equation for m
m2 + 2 = 0 .
This has complex conjugate roots, m = ±i, and so the general solution of (2) is
X(x) = A cos(x) + B sin(x) .
Note that this form of solution (obtained with a negative constant 2) agrees with our physical
intuition about vibrating wires. (What would we get if we had chosen +2?)
There are 3 unknown constants in this general solution, A, B and .
We now use the boundary conditions to say something about these constants:
The boundary condition at x = 0 is u(0, t) = 0. This requires X(0) = 0, implying A = 0, so that
X(x) = B sin(x) .
The boundary condition at x = ` is u(`, t) = 0. This requires X(`) = B sin(`) = 0.
We cannot set B = 0 since this gives the trivial solution X(x) ⌘ 0.
Instead we can satisfy the boundary condition by choice of : sin(`) = 0 when ` = n⇡, n = 1, 2, . . ..
So there are an infinite number of possible values of ,
n =
n⇡
`
, n = 1, 2, . . . ,
and corresponding solutions to (2)
Xn(x) = Bn sin(nx) , n = 1, 2, . . . .
The Bn are still undefined. To see why this does not matter we now consider solutions to (3).
ENG2011
Di↵erential Equations & Series
– 3 – lecture 8
PDEs I
constant
constant
a
X x emx
X mtmx mffffTXAHNHBsi.nl
nx
m2 N 0
Equation (3)
To solve (3) look for solutions in the form T (t) = emt, giving the auxiliary equation for m
m2 + a22 = 0 .
This has complex conjugate roots, m = ±ia, and so the general solution of (3) is
T (t) = C cos(at) + D sin(at) .
But we know that is restricted to a set of values; = n = n⇡/`.
Therefore, for each n, we have a corresponding solution to (3),
Tn(t) = Cn cos(ant) + Dn sin(ant) .
Combining Xn(x) with Tn(t) we obtained a solution un(x, t) to (1)
un(x, t) = Xn(x)Tn(t) = Bn sin(nx)
✓
Cn cos(ant) + Dn sin(ant)
◆
= sin(nx)
✓
↵n cos(ant) + n sin(ant)
◆
with arbitrary constants ↵n = BnCn and n = BnDn.
Since the PDE (1) is linear we can superpose (add) these solutions un to obtain a general solution
u(x, t) =
1X
n=1
un =
1X
n=1
Xn Tn =
1X
n=1
sin(nx)
✓
↵n cos(ant) + n sin(ant)
◆
Finally, the constants ↵n, n need to be determined using the initial conditions. These require
u(x, 0) =
1X
n=1
↵n sin(nx) = 0(x) ,
@u
@t
(x, 0) =
1X
n=1
ann sin(nx) = 1(x) .
The functions 0 and 1 are known.
If we represent these in terms of their Half-Range Fourier Sine Series on [0, `] then we can identify
the required values of ↵n and n to satisfy the initial conditions.
In some cases the answers are immediately obvious. For example, supppose
0 = 3 sin
⇣⇡x
`
⌘
simple sine form initial displacement,
1 = 0 zero initial velocity.
From these we see that ↵1 = 3, ↵n = 0 (n 6= 1), and n = 0 (all n), and the corresponding solution
to equation (1) is
u(x, t) = 3 sin
⇣⇡x
`
⌘
cos
✓
a
⇡t
`
◆
.
ENG2011
Di↵erential Equations & Series
– 4 – lecture 8
PDEs I
The One-Dimensional Heat Equation
The temperature ✓(x, t) along a solid bar of length ` and with an insulated surface, can be modelled
using the one-dimensional heat conduction equation
@✓
@t
= 
@2✓
@x2
, 0 6 x 6 ` , t > 0 , (4)
with di↵usion coecient  determined by  = k/⇢c, k coecient of thermal conductivity, ⇢ material
density, c specific heat capacity.
boundary conditions
There are various thermal constraints we can impose at the ends of the bar, at x = 0 and x = `.
For example, suppose that at x = 0 the bar insulated. This translates into the boundary condition
@✓
@x
(0, t) = 0 , t > 0 .
And suppose that at x = ` the bar is held at a constant temperature ✓c.
Note that the shifted temperature ✓(x, t) ✓c also satisfies equation (4).
Therefore, without loss of generality we can take ✓c = 0 giving the boundary condition
✓(`, t) = 0 , t > 0 .
initial condition
We assume that the initial temperature along the bar is known, or can be prescribed.
This gives the initial condition
✓(x, 0) = (x) , 0 6 x 6 ` ,
where is a given function.
Now look for solutions to (4) in a Separated Variable form
✓(x, t) = X(x)T (t) .
Then
@✓
@t
= X(x) T˙ (t) and
@2✓
@x2
= X 00(x)T (t) .
Substitute these into (4)
X T˙ = X 00 T .
and rearrange (separate variables) to get
1

T˙
T
=
X 00
X
.
As before, the left-hand side is a function of t only, while the right-hand side is a function of x only.
Therefore
1

T˙
T
=
X 00
X
= 2 ,
where 2 is a constant. (As before we will see that the choice of a negative constant is appropriate.)
ENG2011
Di↵erential Equations & Series
– 5 – lecture 8
PDEs I
The two ODEs, for X and T , are now
X 00 + 2X = 0 , (5)
T˙ + 2T = 0 . (6)
Equation (5)
To solve (5) look for solutions in the form X(x) = emx, giving the auxiliary equation for m
m2 + 2 = 0 .
This has complex conjugate roots, m = ±i, and so the general solution of (5) is
X(x) = A cos(x) + B sin(x) .
Now use the boundary conditions to say something about A, B and :
The boundary condition at x = 0 is @✓/@x = 0. This requires X 0(0) = 0, implying B = 0, so that
X(x) = A cos(x) .
The boundary condition at x = ` is ✓(`, t) = 0. This requires X(`) = A cos(`) = 0.
We cannot set A = 0 since this will give X(x) ⌘ 0.
Instead we choose such cos(`) = 0. This requires ` = ⇡/2 + n⇡, n = 0, 1, 2, . . ..
So there are an infinite number of possible values of ,
n =
(2n+ 1)⇡
2`
, n = 0, 1, 2, . . . ,
and corresponding solutions to (5)
Xn(x) = An cos(nx) , n = 0, 1, 2, . . . .
Now consider solutions to (6).
Equation (6)
To solve (6) look for solutions in the form T (t) = emt. The auxiliary equation for m is
m + 2 = 0 ,
giving m = 2, and the general solution
T (t) = C exp(2t) .
We know that is restricted to a set of values; = n = (2n+ 1)⇡/2`. Therefore, for each n, we
have a corresponding solution to (3),
Tn(t) = Cn exp(2nt) .
Combining Xn(x) with Tn(t) we obtained a solution ✓n(x, t) to (4)
✓n(x, t) = Xn(x)Tn(t) = ↵n cos(nx) exp(2nt)
with arbitrary constant ↵n = AnCn.
ENG2011
Di↵erential Equations & Series
– 6 – lecture 8
PDEs I
Since the PDE (4) is linear we superpose these solutions ✓n to obtain a general solution
✓(x, t) =
1X
n=0
✓n =
1X
n=0
Xn Tn =
1X
n=0
↵n cos(nx) exp(2nt) .
The constants ↵n are determined by the initial condition. This requires
✓(x, 0) =
1X
n=0
↵n cos(nx) = (x) .
The function is known, and if this is represented by a suitable Half-Range Fourier Cosine Series
on [0, `] then the required values of ↵n can be determined.
If is given in the form of the cosine series then we can identify the ↵n immediately.
For example, if
(x) = 7 cos
✓
3⇡x
2`
◆
6 cos
✓
5⇡x
2`
◆
then ↵1 = 7, ↵2 = 6, ↵n = 0 (n 6= 1, 2).
In the final lecture we will apply the same method to obtain solutions to Laplace’s Equation.
ENG2011
Di↵erential Equations & Series
– 7 – lecture 8
PDEs I
DIFFERENTIAL EQUATIONS & SERIES
LECTURE 9
PARTIAL DIFFERENTIAL EQUATIONS II
In this lecture we apply the Separation of Variables technique to Laplace’s Equation.
To simplify the illustrations we consider the two dimensional form for (x, y)
r2 = @
2
@x2
+
@2
@y2
= 0 . (⇤)
We might imagine to represent the steady-state temperature in some region of the x-y plane.
We will need to attach boundary conditions to (⇤) at all points on the boundary of this region.
These conditions usually specify the temperature (), or the temperature gradient normal to the
boundary (r ·n), or sometimes a relation between of these two.
We look for solutions to (⇤) is the separated variable form
(x, y) = X(x)Y (y) .
Substitute into (⇤)
X 00Y + XY 00 = 0 ,
and rearrange
X 00
X
= Y
00
Y
.
Note that the left-hand side is a function of x only, and the right-hand side a function of y only.
It follows that neither side can depend on x or y, and they must equal some constant value.
X 00
X
= Y
00
Y
= K constant .
In the previous lecture we chose this constant to have a negative value (K = 2).
Here, the sign will depend on the nature of the boundary conditions (see examples later).
The resulting ODEs for X and Y are
X 00 KX = 0 ,
Y 00 + KY = 0 .
The form of the general solutions will depend on the sign of K.
Suppose K is negative. Write K = 2. Then the general solutions can be written
X(x) = A cos(x) + B sin(x) ,
Y (y) = C exp(y) + D exp(y) .
It is sometimes convenient to rewrite the expression for Y di↵erently, in terms of hyperbolic functions
cosh(z) = 12
ez + ez
, sinh(z) = 12
ez ez .
Using these we can write
Y (y) = C cosh(y) + D sinh(y) , (di↵erent C, D).
It remains to find the values of the constants A, B, C, D, consistent with the boundary conditions.
ENG2011
Di↵erential Equations & Series
– 1 – lecture 9
PDEs II
X N x ikek my
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以 ⼼ 0 以 ⼟⼊
sinhly 主以 ēy
Cosh lol I
sin h o 0
example 1
Find the solution of Laplace’s equation
@2
@x2
+
@2
@y2
= 0
on the rectangular region 0 6 x 6 2, 0 6 y 6 1, satisfying the boundary conditions
(x, 0) = 0 ,
(x, 1) = 0 ,
(0, y) = 0 ,
(2, y) = 5 sin(2⇡y) .
The final boundary condition suggests taking K = +2 so that the general solutions for X and Y take the
form
X(x) = A cosh(x) + B sinh(x) ,
Y (y) = C cos(y) + D sin(y) .
The boundary condition (x, 0) = 0 requires C = 0.
Then the boundary condition (x, 1) = 0 requires = n = n⇡, n = 1, 2, . . ..
Therefore
(x, y) =
1X
n=1
⇣
↵n cosh(nx) + n sinh(nx)
⌘
sin(ny) .
Recall that cosh(0) = 1 and sinh(0) = 0.
The boundary condition (0, y) = 0 therefore requires ↵n = 0 (all n), so that
(x, y) =
1X
n=1
n sinh(nx) sin(ny) .
The final boundary condition then requires
(2, y) =
1X
n=1
n sinh(2n) sin(ny) = 5 sin(2⇡y) .
Since n = n⇡ we can satisfy this condition by taking a single term (n = 2), setting 2 sinh(22) = 5.
The solution is therefore
(x, y) =
5
sinh(4⇡)
sinh(2⇡x) sin(2⇡y) .
ENG2011
Di↵erential Equations & Series
– 2 – lecture 9
PDEs II
i
中州 go.is
0go z
sxkitp
oetssoeontocoeouthose.net
example 2
Find the solution of Laplace’s equation
@2
@x2
+
@2
@y2
= 0
on the rectangular region 0 6 x 6 1, 0 6 y 6 3, satisfying the boundary conditions
@
@y
(x, 0) = 0 ,
(x, 3) = 0 cos
⇣⇡x
2
⌘
,
@
@x
(0, y) = 0 ,
(1, y) = 0 .
The boundary condition at y = 3 suggests taking K = 2 so that the general solutions for X and Y take
the form
X(x) = A cos(x) + B sin(x) ,
Y (y) = C cosh(y) + D sinh(y) .
The boundary condition at x = 0 requires B = 0.
Then the boundary condition at x = 1 requires = n =
1
2⇡ + n⇡, n = 0, 1, 2, . . ..
Therefore
(x, y) =
1X
n=0
cos(nx)
⇣
↵n cosh(ny) + n sinh(ny)
⌘
.
The boundary condition at y = 0 therefore requires n = 0 (all n), so that
(x, y) =
1X
n=0
↵n cos(nx) cosh(ny) .
Finally, the boundary condition at y = 3 requires
1X
n=0
↵n cos(nx) cosh(3n) = 0 cos
⇣⇡x
2
⌘
.
Since n =
1
2⇡+n⇡ we can satisfy this condition by taking a single term (n = 0), setting ↵0 cosh(30) = 0.
The solution is therefore
(x, y) =
0
cosh
3⇡
2
cos⇣⇡x
2
⌘
cosh
⇣⇡y
2
⌘
.
ENG2011
Di↵erential Equations & Series
– 3 – lecture 9
PDEs II
example 3
Find the solution of Laplace’s equation
@2
@x2
+
@2
@y2
= 0
on the rectangular strip 0 6 x 6 1, 0 6 y <1, satisfying the boundary conditions
(x, 0) = 7 cos
⇣⇡x
2
⌘
,
lim
y!1
(x, y) = 0 ,
@
@x
(0, y) = 0 ,
(1, y) = 0 .
The boundary condition at y = 0 suggests taking K = 2.
And the boundary condition as y ! 1 then prompts writing the general solutions for X and Y in the
form
X(x) = A cos(x) + B sin(x) ,
Y (y) = C exp(y) + D exp(y) .
The boundary condition x = 0 requires B = 0.
Then the boundary condition at x = 1 requires = n =
1
2⇡ + n⇡, n = 0, 1, 2, . . ..
Therefore
(x, y) =
1X
n=0
cos(nx)
⇣
↵n exp(ny) + n exp(ny)
⌘
.
The boundary condition as y !1 requires ↵n = 0 (all n), so that
(x, y) =
1X
n=0
n cos(nx) exp(ny) .
Finally, the boundary condition at y = 0 requires
1X
n=0
n cos(nx) = 7 cos
⇣⇡x
2
⌘
.
Since n =
1
2⇡ + n⇡ we can satisfy this condition by taking a single term (n = 0), setting 0 = 7.
The solution is therefore
(x, y) = 7 cos
⇣⇡x
2
⌘
exp
⇣
⇡y
2
⌘
.
ENG2011
Di↵erential Equations & Series
– 4 – lecture 9
PDEs II
y
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