THE UNIVERSITY OF MELBOURNE
SCHOOL OF MATHEMATICS AND STATISTICS
MAST20029 Engineering Mathematics
Lecture Notes
STUDENT NAME:
This compilation has been made in accordance with the provisions of Part VB of the copyright act for the teaching purposes of the University.
This booklet is for the use of students of the University of Melbourne enrolled in the subject MAST20029 Engineering Mathematics.
© School of Mathematics and Statistics, University of Melbourne, February 2022.
These notes have been written by Christine Mangelsdorf and Marcus Brazil at the University of Melbourne. Reproduction of any part of this work other than that authorised by
Australian Copyright Law without permission of the copyright owners is unlawful.
Edition 17, February 2022.
MAST20029 Engineering Maths: Vector Calculus
1.1
VECTOR CALCULUS
MAST20029 Engineering Maths: Vector Calculus
1.2
Vector Fields [Kreyszig, p.375-378]
Physical examples of vector fields are force fields (eg, due to
gravity or electrical charge) and velocity fields (eg, due to air
or fluid flow).
A vector field is a function
Example
F : Rn ! Rn
n=2
F(x, y) = u(x, y)i + v(x, y)j
• 2 independent variables x, y
• F is a 2 component vector (u, v) in 2D where each component is a scalar function of x, y.
A tank of water is stirred so that the water rotates around a
central point. The velocity of the water at any point in the
tank is given by the vector field
v : R3 ! R3 where
v(x, y, z) = yi
xj
v has no k component because the rotation is planar.
Sketching Vector Fields
n=3
F(x, y, z) = u(x, y, z)i + v(x, y, z)j + w(x, y, z)k
• 3 independent variables x, y, z
• F is a 3 component vector (u, v, w) in 3D where each
component is a scalar function of x, y, z.
The vector field can be sketched by assigning to each point
(x, y, z) a vector F represented by an arrow whose tail is at
(x, y, z).
MAST20029 Engineering Maths: Vector Calculus
(1) Using vertical strips:
1.15
MAST20029 Engineering Maths: Vector Calculus
(2) Using horizontal strips:
1.16
MAST20029 Engineering Maths: Vector Calculus
1.21
MAST20029 Engineering Maths: Vector Calculus
1.22
Change of Variables in R2
[Kreyszig, p.429 - 432]
Suppose we have a two-dimensional change of variable
x
=
x(u, v)
y
=
y(u, v)
The Jacobi matrix for this change of variables is
3
2
@x @x
6 @u @v 7
7
6
7
6
4 @y @y 5
@u @v
The determinant of the Jacobi matrix is called the Jacobian
determinant and is denoted by J(u, v).
MAST20029 Engineering Maths: Vector Calculus
1.27
MAST20029 Engineering Maths: Vector Calculus
1.28
MAST20029 Engineering Maths: Vector Calculus
1.35
MAST20029 Engineering Maths: Vector Calculus
1.36
Change of Variables in R3
Suppose we have a three-dimensional change of variable
x
=
x(u, v, w)
y
=
y(u, v, w)
z
=
z(u, v, w)
The Jacobi matrix for this change of variables is
2
3
@x @x @x
6 @u @v @w 7
6
7
6
7
6
7
6 @y @y @y 7
6
7
6 @u @v @w 7
6
7
6
7
4 @z @z @z 5
@u @v @w
The determinant of the Jacobi matrix is called the Jacobian
determinant and is denoted by J(u, v, w).
MAST20029 Engineering Maths: Vector Calculus
1.39
Cylindrical coordinates are an extension of polar coordinates
to 3 dimensions.
MAST20029 Engineering Maths: Vector Calculus
The Jacobian determinant is
• P 0 is the projection of P (r, ✓, z) onto the xy-plane.
J(r, ✓, z)
• r and ✓ are measured for P 0 as in polar coordinates.
=
• z is the z-coordinate of P .
@x
@r
@x
@✓
@x
@z
@y
@r
@y
@✓
@y
@z
@z @z @z
@r @✓ @z
cos ✓
r sin ✓
sin ✓ r cos ✓
0
0
• x2 + y 2 = r 2 .
=
cos ✓
sin ✓
=
r sin ✓
r cos ✓
=
r cos2 ✓ + r sin2 ✓
=
r
The integral becomes
ZZZ
ZZZ
f (x, y, z)dxdydz =
V
0
0
1
V⇤
where F (r, ✓, z) = f (r cos ✓, r sin ✓, z).
F (r, ✓, z)r drd✓dz
1.40
MAST20029 Engineering Maths: Vector Calculus
1.41
Exercise 12
Find the volume of the solid region V which lies inside the
cylinder x2 + y 2 = 1, z 2 R, below the plane z = 4 and above
the paraboloid z = 1 x2 y 2 .
Solution
Using cylindrical coordinates, V is described as
MAST20029 Engineering Maths: Vector Calculus
1.42
MAST20029 Engineering Maths: Vector Calculus
1.45
The Jacobian determinant J(r, , ✓) is
=
=
@x
@r
@x
@
@x
@✓
@y
@r
@y
@
@y
@✓
@z
@r
@z
@
@z
@✓
cos ✓ sin
sin ✓ sin
cos
MAST20029 Engineering Maths: Vector Calculus
The integral becomes
ZZZ
ZZZ
f (x, y, z)dxdydz =
V
V
F (r, , ✓)r2 sin drd d✓
⇤
where F (r, , ✓) = f (r cos ✓ sin , r sin ✓ sin , r cos ).
r cos ✓ cos
r sin ✓ cos
r sin
r sin ✓ sin
r cos ✓ sin
0
=
cos
r cos ✓ cos
r sin ✓ cos
r sin ✓ sin
r cos ✓ sin
+
r sin
cos ✓ sin
sin ✓ sin
r sin ✓ sin
r cos ✓ sin
=
cos (r2 cos2 ✓ sin cos + r2 sin2 ✓ sin cos )
+
r sin (r cos2 ✓ sin2
=
r2 sin cos2
=
r2 sin
+ r sin2 ✓ sin2 )
+ r2 sin3
1.46
MAST20029 Engineering Maths: Vector Calculus
1.57
MAST20029 Engineering Maths: Vector Calculus
1.58
MAST20029 Engineering Maths: Vector Calculus
Solution
1.63
MAST20029 Engineering Maths: Vector Calculus
1.64
Conservative Fields
[Kreyszig, p.421-424]
A work integral is path independent if
Z
Z
F · dr = F · dr
C1
C2
for any two simple oriented curves C1 , C2 with the same
endpoints.
Let
F(x, y, z) = F1 i + F2 j + F3 k
be a vector field such that F1 , F2 and F3 have continuous
first-order partial derivatives.
R
We say that F is a conservative vector field if C F · dr is path
independent. This is equivalent to
Z
F · dr = 0 for any simple closed curve C.
1.
C
2. r ⇥ F = 0
3. F = r for some scalar function
MAST20029 Engineering Maths: Vector Calculus
1.65
Theorem
Let C be a path starting at the point A and finishing at the
R
point B. If F is conservative and F = r , then C F · dr
depends only on the endpoints of C and
Z
F · dr = (B)
(A)
C
MAST20029 Engineering Maths: Vector Calculus
Exercise 21
R
Let F(x, y) = xi + yj. Evaluate C F · dr along
(a) y = x2 from (0, 0) to (1, 1);
(b) line segments joining (0, 0) to (0, 1) to (1, 1);
(c) if C is the unit circle.
Solution
1.66
MAST20029 Engineering Maths: Vector Calculus
1.67
MAST20029 Engineering Maths: Vector Calculus
1.68
Exercise 22
Consider the vector field F(x, y, z) = (x2 , cos y sin z, sin y cos z).
R
Evaluate C F · dr where
r(t) = (t2 + 1, et , e2t ), 0  t  1
Solution
MAST20029 Engineering Maths: Vector Calculus
1.69
MAST20029 Engineering Maths: Vector Calculus
1.70
Surface Integrals
[Kreyszig, p.448-450]
Let S be a smooth surface defined by z = f (x, y). We define
the surface integral of g over S to be
ZZ
S
g(x, y, z)dS =
ZZ
R
q
g(x, y, f (x, y)) fx2 + fy2 + 1dydx
where R is the region formed by projecting S onto the xyplane.
By projecting vertically, we turn the surface integral of g over
S into a double integral over the region R in the xy-plane.
Note
In this subject we only consider surface integrals for surfaces
that can be written in the form z = f (x, y).
MAST20029 Engineering Maths: Vector Calculus
1.75
MAST20029 Engineering Maths: Vector Calculus
1.76
Flux Across a Surface
[Kreyszig, p.443-447]
Let S be a smooth oriented surface defined by z = f (x, y).
Assume the orientation of S is such that the unit normal vector
n̂ on S is upward. Let
F(x, y, z) = F1 i + F2 j + F3 k
be a vector field.
We define the flux integral of F over S to be
ZZ
S
F · n̂ dS =
ZZ
F1 fx
F2 fy + F3 dydx
R
where R is the region formed by projecting S onto the xyplane.
Note
In this subject we only consider flux integrals for surfaces that
can be written in the form z = f (x, y).
MAST20029 Engineering Maths: Vector Calculus
1.83
MAST20029 Engineering Maths: Vector Calculus
1.84
MAST20029 Engineering Maths: Vector Calculus
1.87
MAST20029 Engineering Maths: Vector Calculus
Exercise 27
Let F(x, y, z) = z 2 k. Let S be the unit sphere
x2 + y 2 + z 2 = 1
Assume S is oriented using the outward unit normal.
Use Gauss’ theorem to find the flux of F across S.
Solution
1.88
MAST20029 Engineering Maths: Vector Calculus
1.89
MAST20029 Engineering Maths: Vector Calculus
1.90
Stokes’ Theorem
[Kreyszig, p.463-468]
Let S be an oriented smooth open surface bounded by the
curve C. Let C be oriented by the right-hand rule, i.e. anticlockwise if the unit normal n̂ on S is upwards, and clockwise
if n̂ is downwards.
Let
F(x, y, z) = F1 i + F2 j + F3 k
be a vector field such that F1 , F2 and F3 have continuous
first-order partial derivatives on S.
Then
Z
C
F · dr =
ZZ
S
(r ⇥ F) · n̂ dS
Note
If F is the velocity field of a fluid,
Z
C
F · dr is the circulation
of F. It measures the extent to which the corresponding fluid
motion is a rotation around C.
MAST20029 Engineering Maths: Systems of ODEs
2.1
SYSTEMS OF ORDINARY DIFFERENTIAL
EQUATIONS
MAST20029 Engineering Maths: Systems of ODEs
Solutions of homogeneous constant coefficient systems
To solve the system, try
[Kreyszig, Ch 4, p124-141]
x = we
Coupled first order ODEs arise in models of physical systems
such as chemical reactions that involve tanks of fluids with
di↵erent concentrations and electric circuits with more than
one loop.
=
dx2
dt
=
a11 x1 (t) + a12 x2 (t)
a21 x1 (t) + a22 x2 (t)
This can be written in matrix form as
ẋ = Ax
or
"
dx1
dt
dx2
dt
#
=
"
a11
a21
where A is a constant matrix.
a12
a22
#"
x1
x2
#
t
where w is a constant vector. Then
ẋ = we
t
Substituting into ẋ = Ax gives
A homogeneous constant coefficient system of two first order
ODEs has the form
dx1
dt
2.2
we
Since e
t
6= 0 then
t
= Awe
t
Aw = w
Hence is an eigenvalue of the matrix A and w is the corresponding eigenvector.
If A has 2 linearly independent eigenvectors {w1 , w2 } with
corresponding eigenvalues { 1 , 2 }, then the general solution
to the system is
x(t) = ↵1 w1 e
1t
where ↵1 and ↵2 are constants.
+ ↵ 2 w2 e
2t
MAST20029 Engineering Maths: Systems of ODEs
Exercise 1
Solve the system
dx
dt
dy
dt
=
x + 3y
=
2x + 2y
subject to the initial conditions
x(0) = 0,
Solution
y(0) = 1
2.3
MAST20029 Engineering Maths: Systems of ODEs
2.4
MAST20029 Engineering Maths: Systems of ODEs
2.5
MAST20029 Engineering Maths: Systems of ODEs
2.6
Real, repeated eigenvalues
Suppose A has one repeated eigenvalue (algebraic multiplicity m = 2) and only one linearly independent eigenvector w
(geometric multiplicity g = 1). One solution is:
x = we
t
A second linearly independent solution is
x = (wt + u)e
t
Substituting into ẋ = Ax gives:
( tw + u + w)e
Since e
t
t
= A(wt + u)e
6= 0
tw + u + w = tAw + Au
Since Aw = w, u satisfies
(A
Au
=
I)u
=
u+w
w
The general solution of the system is
x = ↵1 we
t
+ ↵2 (wt + u)e
where ↵1 and ↵2 are constants.
t
t
MAST20029 Engineering Maths: Systems of ODEs
Exercise 2
Solve the system
"
Solution
dx
dt
dy
dt
#
=
"
3
1
1
1
#"
x
y
#
2.7
MAST20029 Engineering Maths: Systems of ODEs
2.8
MAST20029 Engineering Maths: Systems of ODEs
2.9
MAST20029 Engineering Maths: Systems of ODEs
Complex eigenvalues
Exercise 3
If A has two complex conjugate eigenvectors {w1 , w2 } with
corresponding complex conjugate eigenvalues { 1 , 2 }, then
the complex general solution to the system is
Solve the system
"
#
x(t) = C1 w1 e
1t
+ C 2 w2 e
2t
Solution
where C1 and C2 are complex constants.
The real-valued solutions to the system may be obtained from
⇥
x(t) = ↵1 Re w1 e
1t
⇤
⇥
+ ↵2 Im w1 e
where ↵1 and ↵2 are real constants.
dx
dt
dy
dt
1t
⇤
=
"
0
2
1
2
#"
x
y
#
2.10
MAST20029 Engineering Maths: Systems of ODEs
2.11
MAST20029 Engineering Maths: Systems of ODEs
2.12
Phase portraits of linear systems
[Kreyszig, Ch 4, p 141-151]
Consider the linear system
"
# "
dx
a11
dt
=
dy
a21
dt
a12
a22
#"
x
y
#
• The equations define a vector field or a phase plane. At
any point x, we have a velocity vector ẋ = Ax.
• Each solution x(t) = (x(t), y(t)) defines a curve in the
xy-plane called an orbit. The functions x(t) and y(t)
are the parametric equations of the curve.
• At any point on a solution curve, the velocity vector ẋ
is tangential to the curve.
• Arrows on the solution curve indicate the direction of t
increasing.
• The phase portrait is determined by the eigenvalues
1 , 2 and eigenvectors w1 , w2 of the matrix A.
MAST20029 Engineering Maths: Systems of ODEs
2.13
Critical points
MAST20029 Engineering Maths: Systems of ODEs
Case 1:
• Any point (x0 , y0 ) where
dx
=0
dt
1
<
2
2.14
<0
Exercise 4
and
dy
=0
dt
Classify the critical point at the origin and sketch the phase
portrait for the system:
dx
dt
dy
dt
is called a critical point.
• The origin (0, 0) is a critical point of every two dimensional homogeneous linear system.
• P is an asymptotically stable critical point, if all of the
orbits approach P .
• P is a stable critical point, if all of the orbits stay near
P but do not approach P .
• P is an unstable critical point if some or all of the orbits
move away from P .
=
x
=
2y
Solution
The general solution is
"
#
" #
x
0
= ↵1
e
y
1
2t
+ ↵2
"
1
0
#
e
t
MAST20029 Engineering Maths: Systems of ODEs
Case 2:
1
<0<
2.17
2
Exercise 5
Classify the critical point at the origin and sketch the phase
portrait for the system:
dx
dt
dy
dt
=
2x
=
x
4y
3y
Solution
The general solution is
"
#
" #
x
1
= ↵1
e
y
1
2t
+ ↵2
"
4
1
#
et
MAST20029 Engineering Maths: Systems of ODEs
2.18
MAST20029 Engineering Maths: Systems of ODEs
2.21
MAST20029 Engineering Maths: Systems of ODEs
Case 4:
1
=
2
2.22
(m = 2, g = 1)
Exercise 7
Classify the critical point at the origin and sketch the phase
portrait for the system:
dx
dt
dy
dt
=
4x
=
x
y
2y
Solution
The general solution is
"
x
y
#
=
↵1
"
1
1
#
e
3t
+ ↵2
"
t
t
1
#
e
3t
MAST20029 Engineering Maths: Systems of ODEs
Case 6:
2.27
=± i
Exercise 9
Classify the critical point at the origin and sketch the phase
portrait for the system:
dx
dt
dy
dt
=
y
=
4x
cos(2t)
2 sin(2t)
#
Solution
The general solution is
"
#
"
x
= ↵1
y
+ ↵2
"
sin(2t)
2 cos(2t)
#
MAST20029 Engineering Maths: Systems of ODEs
2.28
MAST20029 Engineering Maths: Systems of ODEs
2.31
Autonomous nonlinear systems
MAST20029 Engineering Maths: Systems of ODEs
where
[Kreyszig, Ch 4, p 152-156]
a=
@f
(x0 , y0 ),
@x
b=
@f
(x0 , y0 )
@y
c=
@g
(x0 , y0 ),
@x
d=
@g
(x0 , y0 )
@y
Consider the nonlinear system where RHS has no t dependence
dx
dt
dy
dt
=
f (x, y)
=
g(x, y)
In matrix form, the linearised system becomes
where f (x, y) and g(x, y) have continuous first order partial
derivatives and are not simply linear combinations of x and y.
Let (x0 , y0 ) be a critical point of the non linear system, and
set
X=x
x0 ,
Y =y
y0
The linearisation of the system about the critical point (x0 , y0 )
is given by the linear system
dX
dt
dY
dt
=
aX + bY
=
cX + dY
Ẋ = AX
where
2
@f
6 @x
A = 4 @g
@x
3
@f
@y 7
@g 5
@y
is the Jacobi matrix of f and g, evaluated at (x0 , y0 ).
2.32
MAST20029 Engineering Maths: Systems of ODEs
2.33
Exercise 10
Consider the following nonlinear system
dx
dt
dy
dt
=1
=
x
2xy
y
= f (x, y)
= g(x, y)
Find all critical points of this system, and linearise the system
about each of its critical points.
Solution
MAST20029 Engineering Maths: Systems of ODEs
2.34
MAST20029 Engineering Maths: Systems of ODEs
2.35
MAST20029 Engineering Maths: Systems of ODEs
2.36
Phase portraits of nonlinear systems
Linearisation Theorem
Let (x0 , y0 ) be an isolated critical point of a two dimensional
autonomous nonlinear system. Let
Ẋ = AX
be the linearisation of the system about (x0 , y0 ).
If the eigenvalues of A are such that
1.
1
6=
2,
and
2. neither eigenvalue has Re( ) = 0,
then the orbits of the nonlinear system close to (x0 , y0 ) can
be approximated by the orbits of the linearised system close
to the origin.
MAST20029 Engineering Maths: Laplace Transforms
3.1
LAPLACE TRANSFORMS
MAST20029 Engineering Maths: Laplace Transforms
Laplace transforms
f (t)
[Kreyszig, Chapter 6, pp 203-253]
Let f be a function of a real variable t
0. The Laplace
transform of f , denoted by L[f ], is the function F (s)
L[f (t)] = F (s) =
Z
1
f (t)e
st
dt
F (s) =
Z
1
f (t)e
st
dt
0
1
s
1.
1
2.
tn
3.
eat
4.
sin(at)
5.
cos(at)
6.
sinh(at)
7.
cosh(at)
8.
f 0 (t)
sL[f ]
9.
f 00 (t)
s2 L[f ]
n!
sn+1
0
where s is a complex variable.
A Laplace transform is well defined if
1
s
a
s 2 + a2
1. f is piecewise continuous,
2. In the limit as t ! 1, |f |  Kect , for some finite
constants K, c.
For many functions, the integral only converges for a limited
range of s.
Applications of Laplace transforms include problems involving
heat conduction, fluid flow, chemical reactions and electrical
circuits.
a
s2
s
+ a2
a
s2
a2
s
s2
a2
f (0)
s f (0)
f 0 (0)
3.2
MAST20029 Engineering Maths: Laplace Transforms
3.3
MAST20029 Engineering Maths: Laplace Transforms
Exercise 1
Exercise 2
Using the definition of the Laplace transform, prove that
Using the definition of the Laplace transform, prove that
L[1] =
Solution
1
s
L[t] =
Solution
1
s2
3.4
MAST20029 Engineering Maths: Laplace Transforms
3.5
MAST20029 Engineering Maths: Laplace Transforms
Linearity
Exercise 4
The Laplace transform is linear, so that
Using the definition of the Laplace transform, prove that
L[eat ] =
f (t) = c1 f1 (t) + c2 f2 (t)
then
Solution
L[f ] = c1 L[f1 ] + c2 L[f2 ]
where c1 , c2 2 C.
Exercise 3
Compute the Laplace transform of
f (t) = t2
Solution
2t + 1
1
s
a
, a2C
3.6
MAST20029 Engineering Maths: Laplace Transforms
3.7
MAST20029 Engineering Maths: Laplace Transforms
Exercise 5
Exercise 6
Compute the Laplace transform of
Compute the Laplace transform of
f (t) = cosh(at)
for a 2 C, using the exponential definition of cosh(at).
Solution
f (t) = sin(at)
for a 2 C, using the exponential definition of sin(at).
Solution
3.8
MAST20029 Engineering Maths: Laplace Transforms
3.9
Laplace transforms of derivatives
L[f ] =
Z
1
0
= lim
b!1
0
f (t) e
Z
b
st
f 0 (t) e
h
i
L f (n) = sn L[f ]
⇥
= lim f (b)e
sb
⇤
st
f (0) + s lim
dt
b!1
Z
t=0
b
f (t)e
0
Similarly
=
=
s L[f 0 ]
s2 L[f ]
L[f
t=b
f (0) + sL[f ]
L [f 00 ]
f (0)
sn
2 0
f (0)
···
0
b!1
=
1
f (n
If L[f ] = F (s)
dt
Integration by parts

Z
st
= lim f (t)e
+
sf (t)e
b!1
sn
dt
st
3.10
The Laplace transform of the n-th derivative of f
The Laplace transform of f 0 (t)
0
MAST20029 Engineering Maths: Laplace Transforms
f 0 (0)
s f (0)
f 0 (0)
st
dt
(n)
n
] = s F (s)
n
X1
k=0
sn
1 k (k)
f
(0)
1)
(0)
MAST20029 Engineering Maths: Laplace Transforms
3.11
Inverse Laplace transform
If L[f (t)] = F (s) is the Laplace transform of f , then f is the
inverse Laplace transform of F and is denoted by L 1 [F ]. It
is defined by the complex contour integral
f (t) = L
1
1
[F ] =
2⇡i
Z
MAST20029 Engineering Maths: Laplace Transforms
3.12
Partial Fractions Revision
Let f (x) and g(x) be polynomials, then
f (x)
! degree n
g(x)
! degree d
can be written as the sum of partial fractions if n < d.
c+i1
F (s)e
st
ds
c i1
Denominator
Factor
Note
• This integral is beyond the scope of this subject. We
will use tables to invert Laplace transforms.
(x
a)
(x
a)r
• Linearity also holds for inverse transforms. Hence if
F (s) = c1 F1 (s) + c2 F2 (s)
Partial Fraction Expansion
A
x
a
A1
A2
Ar
+
+ ··· +
x a (x a)2
(x a)r
then
L
1
[F ] = c1 L
where c1 , c2 2 C.
1
[F1 ] + c2 L
1
[F2 ]
Ax + B
+ bx + c
(x2 + bx + c)
(x2 + bx + c)r
x2
A1 x+B1
x2 +bx+c
+
A2 x+B2
(x2 +bx+c)2
+ ··· +
Ar x+Br
(x2 +bx+c)r
MAST20029 Engineering Maths: Laplace Transforms
Exercise 7
MAST20029 Engineering Maths: Laplace Transforms
Using Laplace transforms to solve linear ODEs
Compute the inverse Laplace transform of
F (s) =
Solution
3.13
(s
1
2)(s
Exercise 8
Solve the second order initial value problem
1)
x00
for x(t).
Solution
x = 1,
x(0) = 1,
x0 (0) = 2
3.14
MAST20029 Engineering Maths: Laplace Transforms
3.15
MAST20029 Engineering Maths: Laplace Transforms
3.16
Using Laplace transforms to solve linear systems of 1st
order ODEs
Exercise 9
Solve the system of inhomogeneous di↵erential equations
x0 + y 0 + x + y
y0
x+y
=
=
1
(1)
t
for x(t) and y(t), given that
x(0) = 0 and y(0) = 0.
Solution
(2)
MAST20029 Engineering Maths: Laplace Transforms
3.17
MAST20029 Engineering Maths: Laplace Transforms
3.18
MAST20029 Engineering Maths: Laplace Transforms
3.19
MAST20029 Engineering Maths: Laplace Transforms
The s-shifting theorem
Exercise 10
If L[f (t)] = F (s) then
Find the Laplace transform of
L[e
at
g(t) = t2 e
f (t)] = F (s + a)
Solution
or
L
1
[F (s + a)] = e
at
f (t)
for any real number a.
If s is a real number, the exponential shifts the variable s to
the left by a.
4t
3.20
MAST20029 Engineering Maths: Laplace Transforms
3.21
MAST20029 Engineering Maths: Laplace Transforms
Exercise 11
Exercise 12
Find the inverse Laplace transform of
Find the inverse Laplace transform of
G(s) =
Solution
4s
s2
12
6s + 18
H(s) =
Solution
4s + 12
(s 3)2 + 9
3.22
MAST20029 Engineering Maths: Laplace Transforms
3.25
MAST20029 Engineering Maths: Laplace Transforms
The t-shifting theorem in terms of step functions
Exercise 13
Using step functions, we can write the t-shifting theorem as
Find the inverse Laplace transform of
If L[f (t)] = F (s), then
L[f (t
or
L
1
⇥
e
a)u(t
as
G(s) =
a)] = e
⇤
F (s) = f (t
for any real number a > 0.
as
F (s)
a)u(t
a)
Solution
e
2s
s5
3.26
MAST20029 Engineering Maths: Laplace Transforms
3.27
Exercise 14
Find the Laplace transform of the function
8
>
if 0  t < ⇡
< 2
g(t) =
0
if ⇡  t < 2⇡
>
:
sin t
if t 2⇡
g
2
O
t
⇡
2⇡
3⇡
4⇡
Solution
Write g as a linear combination of regular functions multiplied
by unit step functions.
MAST20029 Engineering Maths: Laplace Transforms
3.28
MAST20029 Engineering Maths: Laplace Transforms
3.31
MAST20029 Engineering Maths: Laplace Transforms
3.32
MAST20029 Engineering Maths: Laplace Transforms
3.33
MAST20029 Engineering Maths: Laplace Transforms
Impulse
The Dirac delta function
Some physical phenomena involve the action of large forces
over short intervals of time, such as a tennis ball hit by a
racquet or a voltage applied to an electric circuit.
Define the Dirac -function for a
For any a
0 and ✏ > 0, let
(
f✏ (t
a) = lim f✏ (t
✏!0
if a  t  a + ✏
otherwise
0
Z
t
(⌧
a) d⌧ = u(t
0
1/✏
Z
0
a a+✏
t
The impulse of a force acting over a short time interval
a  t  a + ✏ is the integral of f✏ from a to a + ✏.
I✏ =
Z
0
1
f✏ (t
a) dt =
0
1
a) =
(
0
1
and
f✏
O
a) =
(
Z
a+✏
a
if t 6= a
if t = a
So that
1
✏
a) =
(t
0 as
1
1
dt = [(a + ✏
✏
✏
a) = 1
1
(⌧
a) d⌧ = 1.
if t < a
if t a
3.34
MAST20029 Engineering Maths: Laplace Transforms
3.35
Laplace transform of Dirac delta function
MAST20029 Engineering Maths: Laplace Transforms
Exercise 16
Solve
L[ (t
a)]
y 00 + 4y 0 + 4y = (t
=
L[lim f✏ (t
=
lim L[f✏ (t a)]
✏!0
Z 1
lim
f✏ (t a)e
✏!0 0
Z a+✏
1 st
lim
e
dt
✏!0 a
✏

t=a+✏
e st
lim
✏!0
s✏
t=a
=
=
=
=
=
=
✏!0
lim
e
as
e
s✏
✏!0
lim
se
✏!0
as
as
s
a)]
e
as
e
1)
with
y(0) = 0,
st
dt
s✏
s✏
by L’Hôpital’s rule
e
Note
Putting a = 0 gives L[ (t)] = 1 and L
1
[1] = (t)
Solution
y 0 (0) = 0
3.36
MAST20029 Engineering Maths: Laplace Transforms
3.37
MAST20029 Engineering Maths: Laplace Transforms
3.38
Laplace transforms of integrals
If f has a Laplace transform L[f ] = F (s), then
L
=
=
=
=
=
=
Z
Z
t
f (⌧ ) d⌧
0
✓Z
1
0
lim
b!1
lim
b!1
lim
b!1
Z

f (⌧ ) d⌧
0
b
0
"
✓Z
◆
e
t
f (⌧ ) d⌧
0
st
◆
e
dt
st
dt
Integration by parts
Z
Z
1
1 st t
e
f (t)e
f (⌧ ) d⌧ +
s
s
0
1
lim
s b!1
F (s)
s
t
1
e
s
Z
sb
Z
b
0
1
f (⌧ ) d⌧ +
s
b
f (t)e
0
st
dt
Z
t=b
st
dt
t=0
b
f (t)e
0
st
dt
#
MAST20029 Engineering Maths: Laplace Transforms
3.39
MAST20029 Engineering Maths: Laplace Transforms
3.40
Exercise 17
Convolution
Solve the integral equation
We define the (Laplace) convolution of two functions f and g
as follows
y(t)
Z
t
y(⌧ ) d⌧ = 3
0
(f ⇤ g)(t) =
Solution
Z
t
f (⌧ )g(t
⌧ ) d⌧
0
The convolution of two functions is symmetric, i.e.
f ⇤g =g⇤f
or
Z
t
f (⌧ )g(t
⌧ ) d⌧ =
0
Z
t
f (t
⌧ )g(⌧ ) d⌧
0
The convolution theorem
If L[f (t)] = F (s) and L[g(t)] = G(s), then
L[(f ⇤ g)(t)] = F (s)G(s)
or
(f ⇤ g)(t) = L
1
[F (s)G(s)]
MAST20029 Engineering Maths: Laplace Transforms
3.41
Exercise 18
Use the convolution theorem to find the inverse Laplace transform of
1
H(s) = 2 2
s (s + 1)
Solution
MAST20029 Engineering Maths: Laplace Transforms
3.42
MAST20029 Engineering Maths: Laplace Transforms
Exercise 19
Solve the integral equation
y(t) =
Solution
1 2
t
2
Z
t
y(⌧ )(t
0
⌧ ) d⌧
3.43
MAST20029 Engineering Maths: Laplace Transforms
3.44
MAST20029 Engineering Maths: Sequences and Series
4.1
SEQUENCES AND SERIES
Sequences
MAST20029 Engineering Maths: Sequences and Series
Standard Limits
1. lim
1
=0
n!1 np
(p > 0)
2. lim rn = 0
(|r| < 1)
A sequence is an ordered list of numbers
a1 , a2 , a3 , a4 . . . a n . . .
n!1
denoted by {an } where an is the nth term (n
1).
A sequence {an } has the limit L (L is finite) if an approaches
L as n approaches infinity. We write:
lim an = L,
Theorem
Let f be a real function and {an } be a sequence of real numbers such that an = f (n). If
x!1
lim an = L
n!1
Note
lim an = L
n!1
6)
lim f (x) = L
x!1
(a > 0)
4. lim n1/n = 1
n!1
5. lim
If the limit exists the sequence converges. Otherwise, the
sequence diverges.
then
n!1
an
=0
n!1 n!
n!1
lim f (x) = L
3. lim a1/n = 1
(a 2 R)
loge n
=0
(p > 0)
np
⇣
a ⌘n
7. lim 1 +
= ea
(a 2 R)
n!1
n
6. lim
n!1
np
=0
n!1 an
8. lim
(p 2 R; a > 1)
4.2
MAST20029 Engineering Maths: Sequences and Series
4.3
Series
S 1 = a1
A series is denoted by
S 2 = a1 + a 2
S 3 = a1 + a 2 + a 3
a n = a1 + a 2 + a 3 + a 4 . . .
..
.
n=1
S n = a1 + a 2 + a 3 + . . . + a n
Example
The sequence {n} = 1, 2, 3, 4, . . ..
1
X
The series
n = 1 + 2 + 3 + 4...
The sequence {Sn } is called the sequence of partial sums.
n=1
If {Sn } converges, that is lim Sn = L, where L is a finite
n!1
The sequence and series diverge to infinity.
value, then we say that the series
1
X
ai converges.
i=1
Example
The sequence
The series
4.4
Given a particular sequence {an }, we can formulate a sequence
of sums,
A series is the sum of all terms in the sequence {an }.
1
X
MAST20029 Engineering Maths: Sequences and Series
⇢
1
10n
If a series does not converge, then it diverges.
= 0.1, 0.01, 0.001, . . ..
1
X
1
= 0.1 + 0.01 + 0.001 + . . . = 0.11111111
n
10
n=1
The sequence converges to 0 while the series converges to
1
.
9
MAST20029 Engineering Maths: Sequences and Series
Exercise 1
4.5
MAST20029 Engineering Maths: Sequences and Series
Exercise 2
th
n
Find the k
Solution
Solution
n=1
⇢
1 1 1
1, , , , . . . .
2 4 8
1 1 1
Evaluate the sum 1 + + + + . . ..
2 4 8
partial sum of the sequence {( 1) }.
1
X
Determine if
( 1)n converges or diverges.
Find the k
th
partial sum of
4.6
MAST20029 Engineering Maths: Sequences and Series
Geometric Series
MAST20029 Engineering Maths: Sequences and Series
The Divergence Test
1
X
an diverges.
If lim an 6= 0 then
n!1
The sum of a geometric series
1
X
4.7
arn = a + ar + ar2 + . . .
Note
n=0
is convergent if |r| < 1 and diverges if |r|
If |r| < 1, we have
1
X
n=0
arn =
1.
a
1
r
Exercise 3
1
1+ +
4
Solution
n!1
✓ ◆2 ✓ ◆ 3
1
1
+
+ ...
4
4
1
X
an may converge or diverge.
n=1
Exercise 4
1
X
n+1
converge or diverge?
Does
n
n=1
Solution
What does the series
converge to?
If lim an = 0 then
n=1
4.8
MAST20029 Engineering Maths: Sequences and Series
4.9
MAST20029 Engineering Maths: Sequences and Series 4.10
Integral Test
Exercise 5
If f is a continuous, positive, decreasing function on [1, 1)
1
X
and an = f (n). Then the series
an
For what values of p > 0 does
n=1
1. converges if
2. diverges if
R1
1
R1
1
f (x) dx converges.
f (x) dx diverges.
Solution
Z
1
1
1
dx converge?
xp
MAST20029 Engineering Maths: Sequences and Series 4.11
MAST20029 Engineering Maths: Sequences and Series 4.12
Harmonic p Series
The harmonic p series
1
X
1
p
n
n=1
converges if p > 1 and diverges if p  1.
Proof
p>0
• f (x) =
•
Z
1
1
1
is continuous, positive and decreasing on [1, 1).
xp
1
dx is convergent if p > 1 and divergent if p  1.
xp
1
X
1
• Hence,
converges if p > 1 and diverges if p  1, by
np
n=1
the integral test.
p0
1
X
1
1
=
6
0,
diverges by the divergence test.
p
n!1 np
n
n=1
• Since lim
Combining cases:
1
X
1
converges if p > 1 and diverges if p  1.
np
n=1
MAST20029 Engineering Maths: Sequences and Series 4.13
MAST20029 Engineering Maths: Sequences and Series 4.14
Exercise 6
1
X
1
Does
converge or diverge?
(n 3)2
n=4
Comparison Test
1
1
X
X
an and
bn be positive term series.
Let
n=1
n=1
Solution
1. If an  bn for all n and
converges.
2. If an
diverges.
1
X
bn converges, then
n=1
bn for all n and
1
X
n=1
1
X
an
1
X
an
n=1
bn diverges, then
n=1
To apply the comparison test we compare a given series to a
harmonic p series or a geometric series.
MAST20029 Engineering Maths: Sequences and Series 4.15
MAST20029 Engineering Maths: Sequences and Series 4.16
Exercise 7
Exercise 8
Determine whether
converges or diverges.
Solution
1
X
n2
n3 + 1
n=1
Determine whether
1
X
n=2
converges or diverges.
Solution
5
n3
1
MAST20029 Engineering Maths: Sequences and Series 4.17
MAST20029 Engineering Maths: Sequences and Series 4.18
Ratio Test
1
X
Let
an be a positive term series and
Exercise 9
1
X
nn
Does
converge or diverge?
n!
n=1
n=1
L = lim
n!1
1. If L < 1,
1
X
an+1
.
an
an converges.
n=1
2. If L > 1,
1
X
an diverges.
n=1
3. If L = 1, the ratio test is inconclusive.
The ratio test is particularly useful if an contains an exponential or factorial function of n.
Solution
MAST20029 Engineering Maths: Sequences and Series 4.19
MAST20029 Engineering Maths: Sequences and Series 4.20
Exercise 10
Alternating Series
Does
converge or diverge?
Solution
1
X
n!n!
(2n)!
n=1
An alternating series has terms which have alternating signs
(+, , +, . . .) or ( , +, , + . . .).
Example
1
X
( 1)n
n
n=1
1
=1
1 1
+
2 3
1 1
+ ...
4 5
Alternating Series Test (Leibniz Test)
If the series
1
X
( 1)n an satisfies
n=1
1. an > 0 for all n
1
2. lim an = 0
n!1
3. an+1  an for all n
N , for some integer N
then the series converges.
If condition 2 is violated, the series diverges.
MAST20029 Engineering Maths: Sequences and Series 4.21
MAST20029 Engineering Maths: Sequences and Series 4.22
Exercise 11
1
X
( 1)n
Does
converge or diverge?
n
n=1
Exercise 12
1
X
3n( 1)n
Does
4n 1
n=1
Solution
Solution
1
converge or diverge?
MAST20029 Engineering Maths: Sequences and Series 4.23
MAST20029 Engineering Maths: Sequences and Series 4.24
Power Series
Theorem
1
X
cn (x a)n are of three types:
A general power series about x = a has the form
n=0
1
X
cn (x
a)n = c0 + c1 (x
a) + c2 (x
a)2 + c3 (x
a)3 . . .
1. convergent only when x = a
n=0
where x 2 R is variable and cn 2 R are constants for n
0.
2. convergent for all x
3. convergent for all |x
|x a| > R
a| < R and divergent for all
R is called the radius of convergence.
• In case (1), R = 0.
• In case (2), R = 1.
The interval of convergence is the interval of x for which the
series converges.
MAST20029 Engineering Maths: Sequences and Series 4.25
MAST20029 Engineering Maths: Sequences and Series 4.26
Generalised Ratio Test
1
X
Let
an and
Exercise 13
n=1
L = lim
n!1
1. If L < 1,
1
X
an+1
.
an
Solution
an converges.
n=1
2. If L > 1,
1
X
Find the radius of convergence and interval of convergence of
1
X
( 3)n xn
.
the series
n2 + 1
n=1
an diverges.
n=1
3. If L = 1, the generalised ratio test is inconclusive.
MAST20029 Engineering Maths: Sequences and Series 4.27
MAST20029 Engineering Maths: Sequences and Series 4.28
MAST20029 Engineering Maths: Sequences and Series 4.29
Exercise 14
Using the generalised ratio test, find the radius of convergence
1
X
(x 1)n
.
and interval of convergence of the series
2n
n=1
Solution
MAST20029 Engineering Maths: Sequences and Series 4.30
MAST20029 Engineering Maths: Sequences and Series 4.33
MAST20029 Engineering Maths: Sequences and Series 4.34
The approximation improves as we take higher order terms. In
general
Exercise 15
f (x) =f (a) + f 0 (a)(x
f 00 (a)(x
2!
a)2
a)+
+
⇡
Find the 4th order Taylor polynomial about x = for
2
f (x) = cos x
Solution
f 000 (a)(x
3!
a)3
+ ...
This is called a Taylor series. If a = 0, then the Taylor series
is called a Maclaurin series.
We can approximate a di↵erentiable function f by a truncated
Taylor series. We call the polynomial approximation a Taylor
polynomial of degree n.
f (x) ⇡ Pn (x) =
n
X
f (k) (a)(x
k=0
k!
a)k
⇣⇡⌘
⇣⇡ ⌘ ⇣
⇡ ⌘ f 00 ⇡2 x
+f
x
+
P4 (x) =f
2
2
2
2!
4
⇡ 3
000 ⇡
iv ⇡
f
x 2
x ⇡2
f
2
2
+
+
3!
4!
0
⇡ 2
2
MAST20029 Engineering Maths: Sequences and Series 4.35
MAST20029 Engineering Maths: Sequences and Series 4.36
Exercise 16
Error in Taylor Polynomials
Find the Maclaurin series for f (x) = loge (1 + x)
Taylor’s Theorem
Solution
The nth order Taylor Polynomial for f around x = a is
Pn (x) = f (a) + f 0 (a)(x
a) + . . . +
f n (a)(x
n!
a)n
The truncation error or remainder for the nth order Taylor
Polynomial is
So,
P (x) = f (0) + f 0 (0)x +
Rn (x) = f (x)
00
2
000
3
iv
Pn (x)
4
f (0)x
f (0)x
f (0)x
+
+
+. . .
2!
3!
4!
Rn (x) =
f n+1 (c)(x a)n+1
(n + 1)!
where c lies between a and x.
If Rn ! 0 as n ! 1 the Taylor series converges to f .
MAST20029 Engineering Maths: Sequences and Series 4.37
Exercise 17
Find an upper bound on the error when loge (1 + x) is approximated by a 3rd order Maclaurin polynomial for |x|  0.1.
Solution
Since loge (1 + x) ⇡ x
and f iv (x) =
the error is
6
(1 + x)4
x3
x2
+
2
3
MAST20029 Engineering Maths: Sequences and Series 4.38
MAST20029 Engineering Maths: Sequences and Series 4.39
Exercise 18
Approximate (1.2)7/2 using a 2nd order Maclaurin polynomial
for (1 + 2x)7/2 . Give an upper bound on the error.
Solution
MAST20029 Engineering Maths: Sequences and Series 4.40
MAST20029 Engineering Maths: Sequences and Series 4.41
Exercise 19
(a) Determine the Maclaurin series for ex and find its radius
of convergence.
1
X
1
= e.
(b) Show that
n!
n=0
Solution
The Maclaurin series is
P (x) = f (0) + f 0 (0)x +
f 000 (0)x3
f 00 (0)x2
+
+ ...
2!
3!
MAST20029 Engineering Maths: Sequences and Series 4.42
MAST20029 Engineering Maths: Fourier Series
5.1
FOURIER SERIES
MAST20029 Engineering Maths: Fourier Series
5.2
Many periodic functions can be expressed as a Fourier series.
Assume that a function f has period T = 2L. Let ! =
[Kreyszig, p.474-494]
A function f is periodic if there is a positive number T such
that
f (t) = f (t + T )
for all t
2⇡
T .
Then f can be represented by a Fourier series of the form:
f (t) = a0 +
1
X
(an cos(n!t) + bn sin(n!t))
n=1
T is called the period of f . The graph of f is obtained by
periodic repetition of its graph in any interval of length T .
Example
The coefficients a0 , an and bn are known as the Fourier coefficients and may be calculated from Euler’s formulae:
a0
=
Example
an
=
Periodic functions describe rotating machines, sound waves or
seasonal phenomena.
f
bn
=
sin t has period 2⇡,
3T
2T
T
cos(2t) has period ⇡
O
T
2T
3T
t
Z L
1
f (t) dt
2L L
Z
1 L
f (t) cos(n!t) dt,
L L
Z
1 L
f (t) sin(n!t) dt,
L L
n = 1, 2, ...
n = 1, 2, ...
MAST20029 Engineering Maths: Fourier Series
5.5
MAST20029 Engineering Maths: Fourier Series
5.6
MAST20029 Engineering Maths: Fourier Series
5.7
MAST20029 Engineering Maths: Fourier Series
5.8
MAST20029 Engineering Maths: Fourier Series
5.9
MAST20029 Engineering Maths: Fourier Series
5.10
Fourier Series Theorem
If a periodic function f with period T = 2L is di↵erentiable
at all except finitely many points in the interval [ L, L] then
the Fourier series with coefficients given by Euler’s formulae
is convergent.
If f is continuous at a point t0 2 [ L, L] then the value of
the Fourier series is f (t0 ).
If f is discontinuous at a point t0 2 [ L, L] then the value
of the Fourier series is the average value of the left and right
limits of f at t0 .
Example
MAST20029 Engineering Maths: Fourier Series
5.13
MAST20029 Engineering Maths: Fourier Series
5.14
Fourier cosine series for even functions
Fourier sine series for odd functions
Suppose f is an even function with period T = 2L.
Suppose f is an odd function with period T = 2L.
Then f (t) cos(n!t) is an even function, and f (t) sin(n!t) is
an odd function.
Then f (t) cos(n!t) is an odd function, and f (t) sin(n!t) is
an even function.
So f may be represented by a Fourier cosine series
So f may be represented by a Fourier sine series
f (t)
where
a0
an
=
=
=
a0 +
1
L
2
L
Z
Z
1
X
an cos(n!t)
n=1
L
f (t)
=
where
L
f (t) cos(n!t) dt
0
bn sin(n!t)
n=1
f (t) dt
0
1
X
bn
=
2
L
Z
L
f (t) sin(n!t) dt
0
MAST20029 Engineering Maths: Fourier Series
5.17
MAST20029 Engineering Maths: Fourier Series
5.18
MAST20029 Engineering Maths: Fourier Series
5.19
Half Range Expansions
Suppose a function f is defined only for the interval 0  t  L.
MAST20029 Engineering Maths: Fourier Series
5.20
We can extend f to an even periodic function fe with period
T = 4. Then fe will have a Fourier cosine series.
fe
The function may be represented in terms of either a cosine
or sine series, by extending its definition to the interval
L  t  L.
3
O
Exercise 3
6
4
2
Consider the function
4
2
t
6
1
f (t) = t2
1
Similarly, we can extend f to an odd periodic function fo with
period T = 4. Then fo will have a Fourier sine series.
for 0 < t < 2. Sketch f .
fo
Solution
3
f
3
1
6
O
1
2
t
4
2
O
1
3
2
4
6
t
MAST20029 Engineering Maths: Fourier Series
5.21
Finding Particular Solutions of Ordinary Di↵erential Equations using Fourier Series
MAST20029 Engineering Maths: Fourier Series
Solution
From Exercise 2, the Fourier series for f is
Exercise 4
f (t) =
Suppose the motion of a constrained object subject to an external force f is given by the di↵erential equation
0
y + 3y = f
where y is the displacement of the body at time t and
(
t + ⇡2 ,
⇡t<0
f (t) =
⇡
0t<⇡
t + 2,
with f (t) = f (t + 2⇡).
1
X
an cos(nt)
n=1
where
an =
(
0,
4
n2 ⇡ ,
n even
n odd
Assume that yp can be represented by a Fourier series with
the same period as f , so that
yp (t) = ↵0 +
1
X
(↵n cos(nt) +
n=1
Di↵erentiating with respect to t gives
Find a particular solution yp to the di↵erential equation using
Fourier series.
5.22
n
sin(nt))
MAST20029 Engineering Maths: Fourier Series
5.23
MAST20029 Engineering Maths: Fourier Series
5.24
MAST20029 Engineering Maths: Fourier Series
5.27
MAST20029 Engineering Maths: Fourier Series
5.28
MAST20029 Engineering Maths: Fourier Series
5.33
MAST20029 Engineering Maths: Fourier Series
5.34
Finding Particular Solutions of Ordinary Di↵erential Equations using Fourier Integrals
Exercise 7
Find a particular solution to the di↵erential equation
y 00 + y 0 + 2y = f
where
f (t) =
(
for |t|  1
for |t| > 1
1
0
Solution
From Exercise 5, the Fourier cosine integral for f is:
2
f (t) =
⇡
Z
1
0
sin(!) cos(!t)
d!
!
Assume that yp can be represented as a Fourier integral:
yp (t) =
Z
1
0
↵(!) cos(!t) + (!) sin(!t) d!
MAST20029 Engineering Maths: Fourier Series
5.35
MAST20029 Engineering Maths: Fourier Series
5.36
MAST20029 Engineering Maths: Fourier Series
5.37
MAST20029 Engineering Maths: Second order PDEs
6.1
SECOND ORDER PDES
[Kreyszig, p.540-552, 558-567]
A partial di↵erential equation (PDE) is an equation involving
one or more partial derivatives of an unknown function .
The dependent variable depends on two or more independent variables which are often time t and one or several space
variables x, y and z, e.g. (x, t), (x, y, z).
Order
The order of a PDE is the order of the highest derivative in
the equation.
Linearity
A PDE is linear if
and its partial derivatives appear only
linearly (to the first power). The independent variables can
appear in any way.
Homogeneity
A PDE is homogeneous if each of its terms contains either
or one of its partial derivatives. Otherwise we call it inhomogeneous.
MAST20029 Engineering Maths: Second order PDEs
6.6
MAST20029 Engineering Maths: Second order PDEs
Exercise 2
Solution
Solve Laplace’s equation
Assume the solution has the form
@2
@2
+
=0
@x2
@y 2
(6)
for {(x, y) : 0 < x < 1, 0 < y < 1} subject to the boundary
conditions
(0, y)
=
0
(1, y)
=
0
(x, 0)
=
0
(x, 1)
=
sin(⇡x) + 2 sin(3⇡x)
(x, y) = X(x)Y (y)
Di↵erentiating twice gives
@2
@x2
=
X (x)Y (y)
@2
@y 2
=
X(x)Y (y)
00
00
Substitute into (6)
y
00
1
= sin(⇡x) + 2 sin(3⇡x)
00
X (x)Y (y) + X(x)Y (y) = 0
00
X (x)
=
X(x)
=0
=0
00
Y (y)
Y (y)
This occurs only if both sides are equal to a constant:
O
=0
1
x
00
X (x)
=
X(x)
00
Y (y)
=
Y (y)
6.7
MAST20029 Engineering Maths: Second order PDEs
6.8
Case 1:
This gives two second order ODEs
00
X (x)
X(x) = 0
00
Y (y) + Y (y) = 0
The boundary conditions can be written as
(0, y) = X(0)Y (y) = 0 so
X(0)
=
0
(1, y) = X(1)Y (y) = 0 so
X(1)
=
0
(x, 0) = X(x)Y (0) = 0
Y (0)
=
0
so
(x, 1) = sin(⇡x) + 2 sin(3⇡x)
Solve the ODEs for the three possible cases of the separation
constant :
1.
>0
2.
=0
3.
<0
MAST20029 Engineering Maths: Second order PDEs
>0
6.9
MAST20029 Engineering Maths: Second order PDEs
Case 2:
=0
6.10
MAST20029 Engineering Maths: Second order PDEs
Case 3:
<0
6.11
MAST20029 Engineering Maths: Second order PDEs
6.12
MAST20029 Engineering Maths: Second order PDEs
6.13
MAST20029 Engineering Maths: Second order PDEs
6.14
MAST20029 Engineering Maths: Second order PDEs
6.15
Exercise 3
Solve the wave equation
@2
@x2
@2
=0
@t2
(7)
for {(x, t) : 0 < x < L, t > 0}, subject to the boundary and
initial conditions
@
(0, t)
@x
@
(L, t)
@x
@
(x, 0)
@t
(x, 0)
=
0
=
0
=
0
=
1 + cos
✓
2⇡x
L
t
@
=0
@x
O
@
=0
@x
@
L
=0
@t
0
1
2⇡x C
A
= 1 + cos B@
L
x
◆
MAST20029 Engineering Maths: Second order PDEs
Solution
6.16
MAST20029 Engineering Maths: Second order PDEs
6.17
This gives two second order ODEs
Assume the solution has the form
00
X (x)
00
T (t)
(x, t) = X(x)T (t)
X(x) = 0
T (t) = 0
Di↵erentiating gives
The boundary and initial conditions can be written as
@2
@x2
=
X (x)T (t)
@2
@t2
=
X(x)T (t)
00
@
(0, t) = X 0 (0)T (t) = 0 so
@x
@
(L, t) = X 0 (L)T (t) = 0 so X 0 (L)
@x
@
(x, 0) = X(x)T 0 (0) = 0 so T 0 (0)
@t
◆
✓
2⇡x
(x, 0) = 1 + cos
L
00
Substitute into (7)
00
X (x)T (t)
=
0
=
0
=
0
00
X(x)T (t) = 0
00
00
X (x)
T (t)
=
X(x)
T (t)
This occurs only if both sides are equal to a constant:
00
X 0 (0)
00
X (x)
T (t)
=
=
X(x)
T (t)
Solve the ODEs for the three possible cases of the separation
constant :
1.
> 0 - this gives only trivial solutions
2.
=0
3.
<0
MAST20029 Engineering Maths: Second order PDEs
Case 2:
=0
6.18
MAST20029 Engineering Maths: Second order PDEs
Case 3:
<0
6.19
MAST20029 Engineering Maths: Second order PDEs
6.20
MAST20029 Engineering Maths: Second order PDEs
6.21
MAST20029 Engineering Maths: Second order PDEs
Solution
6.24
MAST20029 Engineering Maths: Second order PDEs
6.25
This gives two ODEs (one first order and one second order)
Assume the solution has the form
00
X (x)
0
T (t)
(x, t) = X(x)T (t)
X(x) = 0
T (t) = 0
Di↵erentiating gives
The boundary and initial conditions can be written as
@2
@x2
=
X (x)T (t)
@
@t
=
X(x)T (t)
00
0
(0, t) = X(0)T (t) = 0 so
X(0)
=
0
(L, t) = X(L)T (t) = 0 so
X(L)
=
0
(x, 0) = f (x)
Substitute into (8)
Solve the ODEs for the three possible cases of the separation
constant :
00
0
X (x)T (t) = X(x)T (t)
00
0
T (t)
X (x)
=
X(x)
T (t)
This occurs only if both sides are equal to a constant:
00
0
X (x)
T (t)
=
=
X(x)
T (t)
1.
> 0 - this gives only trivial solutions
2.
= 0 - this gives only trivial solutions
3.
<0
MAST20029 Engineering Maths: Second order PDEs
Case 3:
<0
6.26
MAST20029 Engineering Maths: Second order PDEs
6.27
MAST20029 Engineering Maths: Second order PDEs
6.28
Exercise 5
MAST20029 Engineering Maths: Second order PDEs
We now need to compare (x, 0) = f (x) with equation (10).
To do this, represent f (x) as a Fourier sine series.
Solve the heat equation
@
@2
=
2
@x
@t
(9)
We obtain the odd periodic extension (period T = 2L)
for {(x, t) : 0 < x < L, t > 0}, subject to the boundary and
initial conditions
(0, t)
=
0
(L, t)
=
0
(x, 0)
=
(x, 0) = f (x) =
(
for 0  x  L/2
for L/2 < x  L
x
L
x
Solution
bn sin
n=1
bn
=
=
Exercise 4 gives the general solution accounting for the homogeneous boundary conditions.
(x, t) =
1
X
⇣ n⇡x ⌘
L
where the Fourier coefficients are
f (x) =
1
X
n=1
6.29
En exp
✓
n2 ⇡ 2 t
L2
◆
sin
⇣ n⇡x ⌘
L
2
L
8
>
<
>
:
Z
L
f (x) sin
0
0
4L
n2 ⇡ 2
4L
n2 ⇡ 2
⇣ n⇡x ⌘
L
dx
for even n
for n = 1, 5, 9, ...
for n = 3, 7, 11, ...
(11)