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Contents
Notation 5
1 Laplace Transform 7
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 Properties of Laplace Transform . . . . . . . . . . . . . . . . . . . . . 8
1.3 Further Properties of the Laplace transform . . . . . . . . . . . . . . 14
1.4 Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.5 Differentiation of Transform . . . . . . . . . . . . . . . . . . . . . . . 17
2 Vector Calculus 19
2.1 Rectangular Coordinates In 3-Space . . . . . . . . . . . . . . . . . . . 19
2.2 Surfaces In 3 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2.1 Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2.2 Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.2.3 Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2.4 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.2.5 Graphs Of Two-variable Functions . . . . . . . . . . . . . . . 28
2.3 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1
2.3.1 Some Applications Of Double Integrals . . . . . . . . . . . . . 29
2.3.2 Double Integrals over Rectangular Regions . . . . . . . . . . . 29
2.3.3 Double Integrals over Non-rectangular Regions . . . . . . . . . 30
2.3.4 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . 32
2.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.5 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.5.1 Line Integrals of Vector fields . . . . . . . . . . . . . . . . . . 40
2.6 Green’s Theorem in the Plane . . . . . . . . . . . . . . . . . . . . . . 44
2.7 Surface Area and Surface Integrals . . . . . . . . . . . . . . . . . . . 46
2.7.1 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
2.7.2 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 47
2.7.3 Surface Integrals of Vector Fields . . . . . . . . . . . . . . . . 48
2.8 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2.9 Gauss’ Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 52
3 Z Transform 55
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.2 Properties of The Z Transform . . . . . . . . . . . . . . . . . . . . . 58
3.3 More Properties of The Z Transform . . . . . . . . . . . . . . . . . . 59
3.4 The Inverse Z Transform . . . . . . . . . . . . . . . . . . . . . . . . . 60
3.5 Solving Difference Equations . . . . . . . . . . . . . . . . . . . . . . . 61
3.6 A Table Of Z Transform . . . . . . . . . . . . . . . . . . . . . . . . . 63
4 Complex Analysis 65
2
4.1 Revision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.2 Loci and Regions of the Complex Plane . . . . . . . . . . . . . . . . . 68
4.3 Functions of a complex variable . . . . . . . . . . . . . . . . . . . . . 69
4.4 More Elementary Functions . . . . . . . . . . . . . . . . . . . . . . . 73
4.5 Complex Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
4.6 Two Integration Methods . . . . . . . . . . . . . . . . . . . . . . . . 80
5 The Fourier Integral and Fourier Transforms 89
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
5.2 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
5.3 Some Properties of the Fourier transform . . . . . . . . . . . . . . . . 95
5.4 A table of Fourier transforms . . . . . . . . . . . . . . . . . . . . . . 99
6 Partial Differential Equations 101
6.1 Revision : Half-range Expansions . . . . . . . . . . . . . . . . . . . . 101
6.2 Revision : 2nd Order HLDE With Constant Coefficients . . . . . . . 102
6.3 Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . 104
6.4 Solving Partial Differential Equations . . . . . . . . . . . . . . . . . . 106
6.5 Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
6.6 Separation Of Variables Method . . . . . . . . . . . . . . . . . . 109
Bibliography 113
3
4
Notation
∵ denotes “since”, “because of”.
∴ denotes “therefore”, “thus”, “hence”.
∃ denotes “there exists”, “there is a/an”, “there are some”.
6 ∃ denotes “there does not exists”.
∀ denotes “for all”, “for every”, “for each”.
∈ denotes “belongs to”.
/∈ denotes “does not belongs to”.
x ∈ A means “x is an element of the set A”
A ⊆ B means “A is a subset of the set B”
R =the set of real numbers
R2 = {(x, y) | x, y ∈ R} denotes the set of all ordered pairs of real numbers.
R3 = {(x, y, z) | x, y, z ∈ R} denotes the set of all ordered triples of real numbers.
C =the set of all complex numbers.
N = {1, 2, 3, · · · } =the set of all natural numbers (positive integers).
Z = {0,±1,±2, · · · } =the set of all integers.
p ⇒ q means “p implies q”
5
6
Chapter 1
Laplace Transform
1.1 Introduction
Definition 1.1.1. Let f(t) be defined for t ≥ 0. Then∫ ∞
0
e−stf(t) dt = limb→∞
∫ b
0
e−stf(t) dt
is called the Laplace transform of f , provided that the improper integral exists.
Note 1.
(a) Usually we denote the Laplace transform of f by L{f(t)} or F (s).
(b) The domain of the transform F (s) is taken to be all values of s for which the
improper integral exists.
(c) If L{f(t)} = F (s), then f(t) = L−1{F (s)} is called the inverse Laplace trans-
form of F.
7
Example 1.
(a) Let f(t) = 1, t ≥ 0. Then L{f(t)} =1
sfor s > 0. In general, L{a} =
a
s, s > 0
where a is a constant.
(b) Let f(t) = t, t ≥ 0. Then L{f(t)} =1
s2for s > 0. Using mathematical induction,
we can show that L{tn} =n!
sn+1or L−1
{ 1
sn+1
}=
tn
(n)!.
(c) Let f(t) = eat, t ≥ 0. Then L{f(t)} =1
s− afor s > a.
1.2 Properties of Laplace Transform
Theorem 1.2.1 (The Linear Property Of The Laplace Transform). Let f and g be
functions whose Laplace transforms exist, and let a and b be constants. Then
L{af(t) + bg(t)} = aL{f(t)}+ bL{g(t)}.
Note 2. The inverse LT is also linear.
Example 2.
(a) Show that L{sin ωt} =ω
s2 + ω2for s > 0.
(b) Show that L{cos ωt} =s
s2 + ω2for s > 0.
Example 3. Find the inverse LT of F (s) =5s− 10
s2 − 5s.
Answer. f(t) = 2 + 3e5t
Reading Assignment 1. Find the inverse LT of G(s) =5s− 4
s3 + 9s.
Answer. By partial fractions,
G(s) =5s− 4
s(s2 + 9)=
A
s+
Bs + C
s2 + 9= · · · = −4
9
1
s+
5
3
3
s2 + 9+
4
9
s
s2 + 9
Taking inverse Laplace transform,
g(t) = −4
9+
5
3sin(3t) +
4
9cos(3t)
8
Definition 1.2.1 (Sufficient Conditions for Existence of L{f(t)}).
(a) A function f is said to be piecewise continuous on [0,∞) if , in any interval
0 ≤ a ≤ t ≤ b, there are at most a finite number of points tk, k = 1, 2, . . . , n,
at which f has finite discontinuities and is continuous on each open interval
tk−1 < t < tk.
(b) A function f is said to be of exponential order ect if there exist constant c and
positive constants M and T such that |f(t)| ≤ Mect for all t > T.
Notes : This says that f(t) does not grow faster than the exponential function
Mect.
Theorem 1.2.2 (Existence of the Laplace Transform). If f(t) is piecewise continuous
on [0,∞) and of exponential order ect, then the Laplace transform of f(t) exists for
all s > c.
Theorem 1.2.3 (L.T of the derivatives of f).
(a) If f(t) is continuous for all t ≥ 0 and is of exponential order eγt, and if f ′(t)
exists and piecewise continuous on every finite interval in the range t ≥ 0, then
the Laplace transform of f ′(t) exists for all s > γ and
L(f ′) = sL(f)− f(0).
(b) If f(t) is piecewise continuous , then
L(f ′′) = s2L(f)− sf(0)− f ′(0).
(c) Hence, by induction
L(f (n)) = snL(f)− sn−1f(0)− sn−2f ′(0)− · · · − f (n−1)(0).
9
Example 4. Use the Laplace Transform method to solve the initial value prob-
lem
(a) y′(t) + 2y(t) = e−t, y(0) = 2.
(b) y′′(t) + 4y(t) = 5e−t, y(0) = 2, y′(0) = 3.
Answer. (a) y = e−t + e−2t (b) y = 2 sin(2t) + cos(2t) + e−t
Reading Assignment 2.
Using Laplace transforms, solve the IVP : y′′ − 5y′ + 6y = 4, y(0) = −3, y′(0) = 7.
Answer. Apply LT to the DE,
[s2Y − sy(0)− y′(0)]− 5[sY − y(0)] + 6Y =4
s
Solve for Y,
[s2Y + 3s− 7]− 5[sY + 3] + 6Y =4
s
(s2 − 5s + 6)Y = 22− 3s +4
s
Y =22− 3s
(s− 2)(s− 3)+
4
s(s− 2)(s− 3)
Using partial fractions,
Y = · · · = 2
3
1
s− 18
s− 2+
43
3
1
s− 3
Invert using the Laplace Transform table,
y =2
3− 18e2t +
43
3e3t
Theorem 1.2.4 ( LT of Integrals). If f(t) is piecewise continuous and is of exponen-
tial order eγt,then
L{∫ t
0
f(τ)dτ}
=1
sL(f)
for s > 0, s > γ.
Theorem 1.2.5 (First Shifting Theorem(s-shifting). If f has Laplace transform F (s)
when s > γ, then eatf(t) has Laplace transform F (s− a) when s− a > γ. That is,
L{eatf(t)
}= F (s− a) or L−1
{F (s− a)
}= eatf(t).
10
Example 5. Find L{eat cosh bt} if L{cosh bt} =s
s2 − b2.
Answer.s− a
(s− a)2 − b2.
Reading Assignment 3.
Find L{eattn} if L{tn} =n!
sn+1.
Answer. Let f(t) = tn so that F (s) =n!
sn+1.
Then L{eattn} = L{eatf(t)
}= F (s− a) =
n!
(s− a)n+1.
Example 6. Solve y′′ + 2y′ + 10y = 0, y(0) = 2, y′(0) = 1.
Answer. Y = · · · = 3 + 2(s + 1)
(s + 1)2 + 32= · · · =
[ 3
s2 + 32+ 2
s
s2 + 32
]s→s−(−1)
y = [sin(3t) + 2 cos(3t)] e−t
Exercise 1. Find L−1
{2s + 7
s2 − 8s + 20
}.
Answer. 2e4t cos(2t) +15
2e4t sin(2t)
Definition 1.2.2 (The Convolution Integral). The convolution of f(t) and g(t) is
the function
(f ∗ g)(t) =
∫ t
0
f(τ)g(t− τ)dτ.
Remark 1. (f ∗ g)(t) = (g ∗ f)(t) =
∫ t
0
g(τ)f(t− τ)dτ
Example 7. Find the convolution of f(t) = sin t and g(t) = t.
Answer. t− sin t.
Theorem 1.2.6 ( Convolution Theorem).
L{(f ∗ g)(t)} = L{f(t)}L{g(t)} = F (s)G(s) or L−1{F (s)G(s)} = (f ∗ g)(t).
Example 8. Find L{t ∗ sin t}.
11
Answer.1
s2(s2 + 1)
Example 9. Find L−1{ 1
(s + 1)(s + 2)
}by convolution.
Answer. e−t − e−2t
Example 10. Solve the IVP
y′′ + y = f(t), y(0) = 0, y′(0) = 1
where f(t) is a continuous function that has a transform.
Note : Unlike the undetermined coefficients method, the Laplace transform method
can solve this IVP even if f(t) is unknown.
Answer. y(t) = sin t + (sin t) ∗ f(t)
Exercise 2. Find the inverse Laplace transforms of the following functions using
convolution.
(a)8
s2(s2 + 4)(b)
9
(s + 1)(s− 2)2(a)
10s
(s2 + 4)(s2 + 9)
Answer. (a) 2t− sin 2t (b) (3t− 1)e2t + e−t (c) 2 cos(2t)− 2 cos(3t)
Definition 1.2.3. An equation of the type
y(t) = f(t) +
∫ t
0
y(τ)g(t− τ)dτ = f(t) + y(t) ∗ g(t)
where f and g are known functions is called Volterra integral equation
12
Example 11. Solve the given Volterra integral equation:
y(t) = t +
∫ t
0
y(τ) sin(t− τ) dτ.
Answer. t +t3
6
Reading Assignment 4. Solve the integral equation y = 3 + e−t
∫ t
0
eτy(τ)dτ
Answer. y = 3 +
∫ t
0
e−(t−τ)y(τ)dτ = 3 + e−t ∗ y(t)
Apply LT,
Y =3
s+
Y
s + 1
Solve for Y,
s
s + 1Y =
3
s
Y =3(s + 1)
s2=
3
s+
3
s2
Taking the inverse Laplace transform,
y = 3 + 3t
Exercise 3. Solve the solution of the given equation:
(a) y(t) = tet +
∫ t
0
τy(t− τ) dτ.
(b) y′ + 3y + 2
∫ t
0
y(τ) dτ = 2t, y(0) = 3.
Answer. (a) y = −1
8e−t +
1
8et +
3
4tet +
1
4t2et
(b) y = 7e−2t − 5e−t + 1
13
1.3 Further Properties of the Laplace transform
Definition 1.3.1. If a ≥ 0, then the unit step function or Heaviside function
H(t− a) has a jump size 1 at t = a and is defined by
H(t− a) =
{0 , t < a
1 , t > a.
Note that H(t) = 1.
Theorem 1.3.1 (Second Shifting Theorem-t-shifting). If f has Laplace transform
F (s), then the function
f(t− a)H(t− a) =
{0 , t < a
f(t− a) , t > a.
has Laplace transform e−asF (s), i.e.
L{f(t− a)H(t− a)
}= e−asF (s) or L−1
{e−asF (s)
}= f(t− a)H(t− a).
Note 3. It is convenient to think of the effect of H(t− a) as to ”switch on” f(t− a)
at time t = a, or to shift f(t) by an amount a along the t-axis.
Example 12. Consider a pulse height c from t = a to t = b
f(t) = c{H(t− a)−H(t− b)}
i.e. switch on at t = a and off at t = b. Find L{f(t)}.
Answer. F (s) =c
s
(e−as − e−bs
)
Example 13. Find L{g(t)} if g(t) =
0 , 0 ≤ t ≤ 1
t− 1 , 1 < t < 2
1 , t ≥ 2
Answer.1
s2(e−s − e−2s)
14
Example 14. Find the inverse LT of F (s) =e−2s
(s + 1)2.
Answer. (t− 2)e−(t−2)H(t− 2)
Example 15. For an RC electric circuit, the current i(t) satisfies
Ri(t) +1
C
∫ t
0
i(τ)dτ = v(t)
where the resistance is R ohms and the capacitance is C farad and the electromotive
force is v volts.
Assuming the circuit is initially quiescent and that R = 100, C = 0.1, use the Laplace
transform to find the current i(t) given that
v(t) =
{0 , t < 1
10(t− 1) , t > 1
Answer. i(t) = (1− e−0.1(t−1))H(t− 1)
Exercise 4. Let f(t) =
{3 , 0 ≤ t < 4
2t− 5 , t ≥ 4
(a) Express f(t) in terms of the unit step functions.
(b) Find L{f(t)
}.
(c) Obtain the response of the harmonic oscillator
x′′ + x = f(t)
to such a forcing function, given that x = 1 anddx
dt= 0 when t = 0.
Answer. (a) f(t) = 3[1−H(t− 4)] + (2t− 5)H(t− 4) (b) L{f(t)
}=
3
s+ 2e−4s · 1
s2
(c) x(t) = 3− 2 cos t + 2[t− 4− sin(t− 4)]H(t− 4)
Exercise 5. Find L−1
{(2s + 7)e−5s
s2 − 8s + 20
}.
Answer.
[2e4(t−5) cos 2(t− 5) +
15
2e4(t−5) sin 2(t− 5)
]H(t− 5)
15
1.4 Dirac Delta Function
Mechanical systems are often acted upon by an impulsive force (or emf in an electrical
circuit) of large magnitude that acts only for a very short period of time. The function
fk(t) =
{1/k , a ≤ t ≤ a + k
0 , otherwise
could serve as a mathematical model for such a force.
(a) The impulse of such a force is
I =
∫ ∞
−∞fk(t)dt =
∫ a+k
a
1
kdt = 1.
(b) L{fk(t)} =1
k
(e−as
s− e−(a+k)s
s
).
(c) The limit δ(t− a) = limk→0+
fk(t) is called the Dirac delta function.
(d) Note that the Dirac delta function is not a proper function. It is a generalized
function characterized by the two properties
(i) δ(t− a) =
{∞ , t = a
0 , t 6= a
(ii) I =
∫ ∞
−∞δ(t)dt = 1
(e) L{δ(t− a)} = e−as
Example 16. Solve y′′ + 2y′ + 5y = 50t− δ(t− 2), y(0) = −4, y′(0) = 10.
You are given that50
s2(s2 + 2s + 5)=
10
s2− 4
s+
4s− 2
s2 + 2s + 5.
Answer. y = 10t− 4− 1
2e−(t−2) sin(2t− 4)H(t− 2)
Exercise 6. Find the solution of the equation
y′′ + 2y′ + 10y = 6δ(t− 2), y(0) = 3, y′(0) = 0.
Answer. y = e−t sin(3t) + 3e−t cos(3t) + 2e−(t−2) sin 3(t− 2)H(t− 2)
16
1.5 Differentiation of Transform
Theorem 1.5.1 (Differentiation of Transform). L(tf(t)
)= −F ′(s) or L−1
(−F ′(s)
)=
tf(t).
Note : It can be shown by induction that L{tnf(t)
}= (−1)n dn
dsnF (s), n = 1, 2, 3, . . . .
Example 17. Find L{t cos ωt
}.
Answer.s2 − ω2
(s2 + ω2)2
Exercise 7. Find L {t2 sin 3t}.
Answer.18s2 − 54
(s2 + 9)3
Example 18. Find L−1
{ln
s− a
s− b
}.
Answer.1
t(ebt − eat)
Exercise 8. Find g(t) = L−1
{ln
(s
s2 + s− 6
)}.
Answer. g(t) =e−3t + e2t − 1
t
17
Table of Laplace Transforms
f(t) F (s) = L{f(t)}1 1/s
tnn!
sn+1(n = 1, 2, 3, . . .)
eat 1
s− a
tneat n!
(s− a)n+1(n = 1, 2, 3, . . .)
sin ωtω
s2 + ω2
cos ωts
s2 + ω2
f ′ sL{f} − f(0)
f ′′ s2L{f} − sf(0)− f ′(0)
eatf(t) F (s− a)
f(t− a)H(t− a) e−asF (s)
tf(t) −F ′(s)
(f ∗ g)(t) =
∫ t
0
f(τ)g(t− τ)dτ F (s)G(s)
δ(t− a) e−as
18
Chapter 2
Vector Calculus
2.1 Rectangular Coordinates In 3-Space
The three mutually perpendicular coordinate axes (the x-,y- and z-axes) form a 3-
dimensional rectangular or Cartesian coordinate system.
Their point of intersection is called the origin of the coordinate system.
Each pair of coordinate axes determines a plane, called a coordinate plane. The
three coordinate planes are called the xy-plane (with equation z = 0); the xz-plane
(with equation y = 0) and the yz-plane.
The coordinate planes divide 3-space into eight octants.
The first octant is the one for which the three coordinates are positive.
In this rectangular coordinate system, a point P is space can be described by an
ordered triple (x, y, z) where
x = directed distance from P to the yz-plane
y = directed distance from P to the xz-plane
z = directed distance from P to the xy-plane
19
2.2 Surfaces In 3 Dimensions
2.2.1 Planes
The linear equation ax+ by + cz = d ((a, b, c) 6= (0, 0, 0)) represents a plane in space.
Figure 2.1: Plane:2x− 5y + z = 4
20
2.2.2 Spheres
A sphere with center (a, b, c) and radius r is the set of all points (x, y, z) such that
the distance between (x, y, z) and (a, b, c) is r. It can be represented by the (standard)
equation
(x− a)2 + (y − b)2 + (z − c)2 = r2.
Figure 2.2: Sphere :x2 + y2 + z2 = 1
Example 19. Some examples are :
Equation Graph
x2 + y2 + z2 = 1 Sphere with center (0,0,0) and radius 1
(x− 1)2 + (y + 2)2 + (z − 3)2 = 42 Sphere with center (1,-2,3) and radius 4
x2 + 2x + y2 − 4y + z2 − 6z + 5 = 0 Sphere with center (-1,2,3) and radius 3
21
2.2.3 Cylinders
An equation that contains only 2 of the variables x, y, and z represents a cylinder
in space (parallel to the axis of the missing variable).
Remark 2. The cylinder can be obtained by graphing the equation in the coordinate
plane of the 2 variables that appear in the equation and then translating that graph
parallel to the axis of the missing variable.
Example 20. Sketch the graph of x2 + y2 = 1 in R3.
Answer. In R2, this equation represent a circle.
In R3, x2 + y2 = 1 is a circular cylinder: it is made of lines parallel to the z−axis
that passing through the circle x2 + y2 = 1 on the xy−plane.
Figure 2.3: Circular Cylinder :x2 + y2 = 1
22
2.2.4 Quadric Surfaces
The equation of a quadric surface in space is an equation of the form
Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0.
The following are the 6 basic types of quadric surfaces:
(a) Ellipsoidx2
a2+
y2
b2+
z2
c2= 1
(b) Hyperboloid of One Sheetx2
a2+
y2
b2− z2
c2= 1
(c) Hyperboloid of Two Sheetz2
c2− x2
a2− y2
b2= 1
(d) Elliptic Conex2
a2+
y2
b2− z2
c2= 0
(e) Elliptic Paraboloid z =x2
a2+
y2
b2
(f) Hyperbolic Paraboloid z =y2
b2− x2
a2= 1
Remark 3. How to sketch a surface?
The shape of a surface can be obtained by considering the curves of intersection be-
tween the surface and some well-chosen planes. The curve of intersection between the
surface and a plane is called the trace of the surface in the plane.
23
Example 21. Sketch and name the surface z = 4 + x2 + y2.
Figure 2.4: Circular Paraboloid:z = 4 + x2 + y2
24
Figure 2.5: Paraboloid
Figure 2.6: Ellipsoid
25
Figure 2.7: Elliptic Cone:z2 =x2
32+
y2
82
Figure 2.8: Hyperbolic Paraboloid :z = y2 − x2
26
Figure 2.9: Hyperboloid of 1-Sheet:x2 + y2 − z2 = 1
Figure 2.10: Hyperboloid of 2-Sheet:z2 − x2 − y2 = 5
27
2.2.5 Graphs Of Two-variable Functions
Definition 2.2.1. The graph of the 2-variable function f(x, y) is the set of points
(x, y, z) for which z = f(x, y) and (x, y) is in the domain of f.
It is also called the surface z = f(x, y).
Example 22. The graph of the function f(x, y) = x+2y+3 is the plane z = x+2y+3.
Example 23. The graph of the function f(x, y) = 4 − x2 − 4y2 is the paraboloid
z = 4− x2 − 4y2.
Example 24. The graph of the function f(x, y) = y2 is the parabolic cylinder z = y2.
28
2.3 Double Integrals
Definition 2.3.1. The double integral of f over a closed region R is defined as
limn→∞
n∑k=1
f(xk, yk)4xk4yk
if the limit exists. If the limit exists, then f is said to be integrable over R and we
denote this limit as
∫∫R
f(x, y)dA where dA = dxdy or dydx.
2.3.1 Some Applications Of Double Integrals
(a) If f(x, y) ≥ 0 and f is continuous on the rectangle R, then the volume of the
solid that lies above R and under the surface z = f(x, y) is given by
V =
∫∫R
f(x, y)dA.
(b) In particular, if f(x, y) = 1 in R, then
∫∫R
f(x, y)dA = A(R), the area of R.
2.3.2 Double Integrals over Rectangular Regions
Theorem 2.3.1. Let R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d} be a closed rectangular
region. If f is continuous on R. then∫∫R
f(x, y)dA =
∫ d
c
∫ b
a
f(x, y)dx dy =
∫ b
a
∫ d
c
f(x, y)dy dx.
Example 25. Evaluate the iterated integral
∫ 3
0
∫ 2
1
(4x + 6y)dx dy.
Answer. 45
Example 26. Find the volume V of the solid under the plane x + y + z = 4 and over
the rectangular region R : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 in the xy-plane.
Answer. 5
29
2.3.3 Double Integrals over Non-rectangular Regions
Theorem 2.3.2. Let f be continuous in the region R.
(a) If R = {(x, y) : a ≤ x ≤ b, g(x) ≤ y ≤ h(x)}, then∫∫R
f(x, y)dA =
∫ b
a
∫ h(x)
g(x)
f(x, y)dy dx.
(b) If R = {(x, y) : g(y) ≤ x ≤ h(y), c ≤ y ≤ d}, then∫∫R
f(x, y)dA =
∫ d
c
∫ h(y)
g(y)
f(x, y)dx dy.
Example 27. Evaluate
∫ ∫R
6xy dA over the region R enclosed between the curves
y =√
x, 2y = x, x = 2 and x = 4.
Answer. 11
Example 28. Use double integral to find the volume of the tetrahedron bounded by
the coordinate planes and the plane 3x + 6y + 4z = 12.
Answer. 4
Example 29. Find the volume of the solid that lies under the paraboloid z = x2 + y2
and above the region bounded by the x-axis, the y-axis, and the line x + y = 1.
Answer. 1/6
Reading Assignment 5.
Find the volume of the solid that lies under the plane 2x + 2y + z = 18 and above the
triangular region R bounded by the lines y = x, y = 2x, and the line x = 2.
Answer. The volume V =
∫ ∫R
(18 − 2x − 2y)dA =
∫ 2
0
∫ 2x
x
(18 − 2x − 2y)dydx =∫ 2
0
[18y − 2xy − y2
]2x
xdx =
∫ 2
0
(18x− 5x2)dx =
[9x2 − 5
3x3
]2
0
=68
3
30
Example 30. Reversing The Order Of Integration
Sketch the region of integration for the integral∫ π
0
∫ π
x
sin y
ydy dx
and write an equivalent integral with the order of integration reversed. Then evaluate
the integral.
Answer. 2
Reading Assignment 6. Evaluate the double integral I =
∫ 1
0
∫ 1
y2
yex2
dx dy by re-
versing the order of integration.
Answer. The region of integration R = {(x, y) : y2 ≤ x ≤ 1, 0 ≤ y ≤ 1} .
Sketch R. (Do it yourself !)
Describe R in another way : 0 ≤ y ≤√
x, 0 ≤ x ≤ 1.
Then I =
∫ 1
0
∫ √x
0
yex2
dy dx =
∫ 1
0
[y2
2
]√x
0
ex2
dx =
∫ 1
0
x
2ex2
dx =
[ex2
4
]1
0
=e− 1
4
31
2.3.4 Double Integrals in Polar Coordinates
Theorem 2.3.3. To evaluate a double integral using polar coordinates, we use∫∫R
f(x, y)dA =
∫∫G
f(r cos θ, r sin θ)rdrdθ,
where G denotes the region of integration in polar coordinates.
Example 31. Use polar coordinates to evaluate the double integral
I =
∫ 1
−1
∫ √1−y2
−√
1−y2
(x2 + y2)dx dy.
Answer. π/2
Example 32. Use polar coordinates to evaluate the double integral I =
∫ 4
0
∫ √16−x2
0
dy dx
(9 + x2 + y2)3/2.
Answer. π/15
Reading Assignment 7.
Use polar coordinates to evaluate the double integral I =
∫ 1
0
∫ √1−y2
−√
1−y2
6x2ydx dy.
Answer. Let R be the the region of integration{
(x, y) : −√
1− y2 ≤ x ≤√
1− y2, 0 ≤ y ≤ 1}
.
In polar coordinates, R = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ π} .
Using the substitution, x = r cos θ, y = r sin θ, r =√
x2 + y2, dA = rdrdθ,
I =
∫ π
0
∫ 1
0
6(r cos θ)2r sin θrdrdθ =
∫ π
0
∫ 1
0
6r4 cos2 θ sin θdrdθ
=
∫ π
0
[6
5r5
]1
0
cos2 θ sin θdθ =
∫ π
0
6
5cos2 θ sin θdθ =
[− 6
15cos3 θ
]π
0
=4
5
Exercise 9. Let R be the annular region lying between the two circles x2 + y2 = 1
and x2 + y2 = 9. Evaluate the integral
∫ ∫R
(x + 2y2)dA.
Answer.
∫ 2π
0
∫ 3
1
[r cos θ + 2 (r sin θ)2] rdrdθ = · · · = 40π
32
2.4 Triple Integrals
Definition 2.4.1. The triple integral of f over a bounded solid region D is defined
as
limn→∞
n∑k=1
f(xk, yk, zk)4xk4yk4zk
provided the limit exists. If it exists, then f is said to be integrable over D. We will
denote the limit as
∫∫∫D
f(x, y, z)dV where dV = dxdydz or dxdzdy or . . . .
Note 4. The volume of D =
∫∫∫D
dV.
Theorem 2.4.1. Fubini’s Theorem
Let B = {(x, y, z) : a ≤ x ≤ b, c ≤ y ≤ d, g ≤ z ≤ h} be a closed rectangular box.
If f is continuous on R. then∫∫∫B
f(x, y, z)dV =
∫ h
g
∫ d
c
∫ b
a
f(x, y, z)dx dy dz.
Note : The iterated integral on the right can be replaced by any of the five other
iterated integrals by changing the order of integration.
Example 33. Evaluate
∫∫∫B
xy2z3dV over the rectangular box
B : 0 ≤ x ≤ 3,−2 ≤ y ≤ 2, 0 ≤ z ≤ 1.
Answer. 6
Theorem 2.4.2. Let R be a closed region in the xy-plane and let g(x, y) and h(x, y)
be continuous functions such that g(x, y) ≤ h(x, y) for all (x, y) ∈ R. If f is integrable
over the region D = {(x, y, z) : (x, y) ∈ R, g(x, y) ≤ z ≤ h(x, y)}, then∫∫∫D
f(x, y, z)dV =
∫ ∫R
∫ h(x,y)
g(x,y)
f(x, y, z)dzdA.
Note 5. Determining The Limits Of Integration
33
To find the limits for a particular order of integration, it is advisable to first determine
the innermost limits, which may be functions of the outer two variables.
After integrating f with respect to the innermost variable, we end up with a double
integral over the projection of D onto the coordinate plane of the outer two variables.
Then, you can determine the remaining limits of integration by methods used for
double integrals.
Example 34. Using a triple integral to find volume
Find the volume of the solid formed by the intersection of the cylinder y = x2 and the
two planes given by z = 0 and y + z = 4.
Answer. V = 256/15
Example 35. Using a triple integral to find volume
Find the volume of the solid enclosed by the paraboloids z = x2+y2 and z = 18−x2−y2.
Answer. 81π
Exercise 10. Find the volume of the region in the first octant bounded above by the
cylinder z = 1− y2 and lying between the vertical planes x + y = 1 and x + y = 3.
Answer. The volume V =
∫ ∫R
∫ 1−y2
0
dzdA = · · · =4
3where R : 1 − y ≤ x ≤
3− y, 0 ≤ y ≤ 1.
34
2.5 Line Integrals
Definition 2.5.1.
(a) A plane curve can be described by a pair of parametric equations
x = f(t), y = g(t), t ∈ I
where f and g are continuous functions on the interval I. The variable t is a
parameter for the curve.
(b) A space curve can be described by a pair of parametric equations
x = f(t), y = g(t), z = h(t), t ∈ I
where f ,g and h are continuous functions on the interval I. A curve can also be
specified by giving the position vector r(t) of a point P = P (t) = (f(t), g(t), h(t)).
That is,
r(t) =−→OP = (f(t), g(t), h(t)) = f(t)i + g(t)j + h(t)k.
The tip of r(t) traces out the curve as t varies over I.
Example 36. The parametric equations for the line passing through the point (x0, y0, z0)
and parallel to the nonzero vector v = (a, b, c) are
x = x0 + at, y = y0 + bt, z = z0 + ct.
Example 37.
(a) Parametrize the line segment joining the point A(1, 4, 3) and B(−2, 5, 6).
(b) Write down the corresponding vector equation for the line segment.
Answer. (a) x = 1 − 3t, y = 4 + t, z = 3 + 3t, 0 ≤ t ≤ 1 (b) r(t) = (1 − 3t)i + (4 +
t)j + (3 + 3t)k, 0 ≤ t ≤ 1
35
Example 38. A Parametrization Of The Circle x2 + y2 = a2
The following is a parametrization for the circle x2 + y2 = a2 :
x = a cos t, y = a sin t, 0 ≤ t ≤ 2π.
Note that the circle is traced out counterclockwise as t increases from t = 0 to t = 2π.
Example 39. A Parametrization Of The Ellipsex2
a2+
y2
b2= 1
x = a cos t, y = b sin t, 0 ≤ t ≤ 2π.
Note that the ellipse is traced out counterclockwise as t increases from t = 0 to t = 2π.
Example 40. Find parametric equations for the portion of the parabola y = x2 − x
joining A(−1, 2) and B(3, 6), oriented from A to B.
Example 41.
(a) Eliminate the parameter to find the Cartesian equation of the curve with para-
metric equations
x = 3t2, y = 5t + 2, 0 ≤ t ≤ 2.
(b) Sketch the curve represented by the parametric equations. Indicate with an arrow
the direction in which t increases.
Answer. x =3
25(y − 2)2, 2 ≤ y ≤ 12
Exercise 11. Find parametric equations for the line segment joining the points
A(2, 3,−4) and B(−5, 0, 1).
Write down the corresponding vector equation for the line segment.
Draw coordinate axes and sketch the line segment, indicate the direction of increasing
t for your parametrization.
Answer. x = 2− 7t, y = 3− 3t, z = −4 + 5t, 0 ≤ t ≤ 1;
r = (2− 7t)i + (3− 3t)j + (−4 + 5t)k, 0 ≤ t ≤ 1;
The line segment with this parametrization moves from A to B.
36
Exercise 12. Find parametric equations for the portion of the parabola x = y2 joining
A(4,−2) and B(9, 3), oriented from A to B.
Answer. x = t2, y = t, −2 ≤ t ≤ 3.
Definition 2.5.2. Let r(t) be the position vector of a curve C defined on an interval
I. We call C a smooth curve or r(t) a smooth function if r′(t) is continuous and
r′(t) 6= 0 for any value of t ∈ I (except possibly at any endpoints of I).
Note : The tangent vector r′(t) for a smooth curve varies ”continuously” without
abrupt changes in direction as t increases. We can think of a smooth curve as a curve
with no ”sharp corners” (called cusps).
Definition 2.5.3. A curve that is made up of a finite number of smooth curves is
called piecewise smooth.
Definition 2.5.4. The length of a smooth curve C : r(t) =< x(t), y(t), z(t) >
, a ≤ t ≤ b, that is traced exactly once as t increases from a to b is given by
L =
∫ b
a
√(dx
dt
)2
+(dy
dt
)2
+(dz
dt
)2
dt or L =
∫ b
a
‖r′(t)‖dt.
Definition 2.5.5. The arc length parameter s for the curve C with base point
P = P (a) is the function
s(t) =
∫ t
a
√(dx
dt
)2
+(dy
dt
)2
+(dz
dt
)2
dt.
Note : By the Fundamental Theorem of Calculus,
ds
dt=
√(dx
dt
)2
+(dy
dt
)2
+(dz
dt
)2
.
Definition 2.5.6. If f is defined on a smooth curve C : x = x(t), y = y(t) a ≤t ≤ b, then the line integral of f along C is∫
C
f(x, y)ds = limn→∞
n∑k=1
f(xk∗, yk
∗)4sk
if the limit exists.
37
Theorem 2.5.1. If f is a continuous function, then∫C
f(x, y)ds =
∫ b
a
f(x(t), y(t))
√(dx
dt
)2
+(dy
dt
)2
dt.
(a) Recall that ds =
√(dx
dt
)2
+(dy
dt
)2
dt.
(b) If f(x, y) ≥ 0, then
∫C
f(x, y)ds represents the area of the ”curved curtain”
whose base is C and whose height above the point (x, y) is f(x, y).
Example 42. Evaluate the line integral I =
∫C
xds, where C is the curve x = t, y =
t2, 0 ≤ t ≤ 1.
Answer. I =
√125− 1
12
Reading Assignment 8.
Evaluate the line integral
∫C
(2 + x2y)ds, where C is the upper half of the unit circle
x2 + y2 = 9.
Answer. A parametric form for C is x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ π.
Hence, ds =
√(dx
dt
)2
+
(dy
dt
)2
dt =
√(−3 sin t)2 + (3 cos t)2dt = 3dt and∫
C
(2 + x2y)ds =
∫ π
0
(2 + (3 cos t)2(3 sin t)
)3dt = 3
[2t− 9 cos3 t
]π
0= 6π + 54
Theorem 2.5.2. If C is a piecewise-smooth curve; i.e., C is the union of a finite
number of smooth curves C1, C2, . . . , Cn, where the end point of Ci−1 is the initial
point of Ci. Then∫C
f(x, y)ds =
∫C1
f(x, y)ds +
∫C2
f(x, y)ds + · · ·+∫
Cn
f(x, y)ds.
Note 6. The line integrals of f with respect to x and y are defined by∫C
f(x, y)dx = limn→∞
n∑k=1
f(xk∗, yk
∗)4xk and
∫C
f(x, y)dy = limn→∞
n∑k=1
f(xk∗, yk
∗)4yk
38
provide the limits exist.
Note :If x = x(t), y = y(t), a ≤ t ≤ b then dx = x′(t)dt, dy = y′(t)dt and∫C
f(x, y)dx =
∫ b
a
f(x(t), y(t))x′(t)dt,
∫C
f(x, y)dy =
∫ b
a
f(x(t), y(t))y′(t)dt.
Example 43. Evaluate the line integral I =
∫C
xydx + (x − y)dy, where C consists
of the line segments from (0, 0) to (2, 0) and from (2, 0) to (3, 2).
Answer. I = 17/3
Example 44. Integrate f(x, y, z) = xy + y + z over the path
C : r(t) = 2ti + tj + (2− 2t)k, 0 ≤ t ≤ 1.
Answer. 13/2
Reading Assignment 9.
Evaluate the line integral I =
∫C
(2+x−yz)ds, where C is the polygonal path consisting
of the circular arc x2 + y2 = 9, z = 2 from (3, 0, 2) to (0, 3, 2) and the line segment
from (0, 3, 2) to (3, 7, 2).
Answer.
(i) C = C1 + C2 where
C1 : x = 3 cos t, y = 3 sin t, z = 2, 0 ≤ t ≤ π
2C2 : x = 3t, y = 3 + 4t, z = 2, 0 ≤ t ≤ 1.
(ii) I = I1 + I2 where I1 =
∫C1
(2 + x− yz)ds and I2 =
∫C2
(2 + x− yz)ds.
(iii) On C1 : ds =
√(dx
dt
)2
+
(dy
dt
)2
+
(dz
dt
)2
dt =
√(−3 sin t)2 + (3 cos t)2 + (0)2 dt =
3dt,
hence I1 =
∫ π/2
0
(2 + 3 cos t− 2(3 sin t))3dt = 3[2t + 3 sin t + 6 cos t
]π/2
0= 3π − 9
39
(iv) On C2 : ds =
√(3)2 + (4)2 + (0)2 dt = 5dt,
hence I2 =
∫ 1
0
(2 + 3t− 2(3 + 4t))5dt = 5[− 4t− 5
2t2
]1
0= −65
2
(v) I = (3π − 9) + (−65/2) = 3π − 83
2
2.5.1 Line Integrals of Vector fields
Definition 2.5.7.
(a) A vector field is a function F that assigns to each point (x1, x2, . . . , xn) in its
domain D a unique vector F(x1, x2, . . . , xn).
(b) A vector field on R3 is a function of the form
F(x, y, z) = M(x, y, z)i + N(x, y, z)j + P (x, y, z)k.
F is continuous if its component functions M, N and P are continuous ;
differentiable if functions of M, N and P are differentiable, and so on.
(c) If the partial derivatives of M, N and P all exist, then
(i) the curl of F is the vector field defined by
curl F =(∂P
∂y− ∂N
∂z
)i−
(∂P
∂x− ∂M
∂z
)j +
(∂N
∂x− ∂M
∂y
)k.
Remark 4. A physical interpretation of curl F
Let the vector field F represents the velocity field in fluid flow. Particles near
(x, y, z) tend to rotate about the axis that points in the direction of curl F
and ‖curl F‖ is a measure of how quickly the particles move around the axis.
If curl F = 0 at a point P , then the fluid is free from rotations at P and F
is said to be ir-rotational.
(ii) The divergence of F is the function defined by
div F =∂M
∂x+
∂N
∂y+
∂P
∂z.
40
(d) By introducing the differential operator ∇ = i∂
∂x+ j
∂
∂y+ k
∂
∂z, we may write
div F = ∇ · F and
curl F = ∇× F =
∣∣∣∣∣∣∣∣∣i j k∂
∂x
∂
∂y
∂
∂z
M N P
∣∣∣∣∣∣∣∣∣ .
A physical interpretation of div F
Let F be the velocity of a fluid, then div F represents the net rate of change
(with respect to time) of the mass of fluid flowing from the point (x, y, z) per unit
volume. That is, div F measures the tendency of the fluid to diverge form the
point (x, y, z) (div F > 0 ) or accumulate toward (x, y, z) (div F < 0 ).
If div F = 0, then F is said to be incompressible.
Example 45. Sketch the vector field F(x, y) = −1
2yi +
1
2xj.
Answer. F · r = 0 and |F| = 12|r| ⇒ F is tangent to a circle centered at (0, 0) and
has length equal to 12
the radius of that circle.
Example 46. Suppose a spherical object of mass M is centered at the origin. Derive
the formula for the gravitational field of force F(x, y, z) exerted by the mass on an
object of mass M located at a point P (x, y, z) in space. Then sketch this field.
Answer. F = −GMm
|r|3r
Exercise 13. Let F = 2xyi− 5y3zj + 4xzk. Find the divergence and the curl of F.
Answer. 2y − 15y2z + 4x; 5y3i− 4zj− 2xk
Exercise 14. Let F = 2yzi− 5y3xj + 4yz2k. Find the divergence and the curl of F.
Answer. −15y2x + 8zy; 4z2i + 2yj− (5y3 + 2z)k
41
Definition 2.5.8. Let C : x = x(t), y = y(t), z = z(t), a ≤ t ≤ b be a smooth
curve with unit tangent vector T. Then the work done by a force F in moving a
particle along C from t = a to t = b is
W =
∫C
F ·T ds.
Remarks Different ways to write the work integral:
(a) W =
∫ b
a
F(r(t)) · r′(t)dt =
∫C
F · dr
(b) W =
∫C
Mdx + Ndy + Pdz if F(x, y, z) = M(x, y, z)i+ N(x, y, z)j+ P (x, y, z)k.
Example 47. Find the work done by the force field F(x, y) = x2i− xyj in moving a
particle counterclockwise along the quarter-circle r(t) = cos ti + sin tj, 0 ≤ t ≤ π/2.
Answer. −2
3
Example 48. Find the work done by the force field F(x, y, z) = yi+zj−xk in moving
a particle along the twisted cubic
C : x = t, y = t2, z = t3
from (0, 0, 0) to (1, 1, 1).
Answer. − 1
60
42
Reading Assignment 10. Calculate the work done by F(x, y, z) = i − yj + xyzk
in moving a particle from (0, 0, 0) to (1,−1, 1) along the curve of intersection of the
cylinder y = −x2 and the plane z = x.
Answer. Let C be the curve of intersection.
Method 1
We can describe C using the parametric equations x = t, y = −t2, z = t.
At (0, 0, 0) : t = x = 0 while at (1,−1, 1) : t = x = 1.
On C :
F = i + t2j− t4k
dr
dt=
dx
dti +
dy
dtj +
dz
dtk = i− 2tj + k
F · dr
dt= 1− 2t3 − t4
The work done =
∫C
F · dr =
∫ 1
0
F · dr
dtdt =
∫ 1
0
(1− 2t3 − t4)dt = · · · = 3
10
Method 2
y = −x2, z = x ⇒ dy = −2xdx, dz = dx
The work done =
∫C
F · dr =
∫C
dx− ydy + xyzdz
=
∫ 1
0
dx− (−x2)(−2xdx) + x(−x2)(x)dx =
∫ 1
0
(1− 2x3 − x4)dx = · · · = 3
10
Exercise 15. Find the work done by the force F = −3xzi + 2yj + 4k in the displace-
ment along the line x = −2z, y = 3z from (4,−6,−2) to (0, 0, 0).
Answer. −60
43
2.6 Green’s Theorem in the Plane
Definition 2.6.1. A plane curve C : r = r(t) (a ≤ t ≤ b) has an initial end point
at r(a) and a final end point at r(b). C is said to be simple if it does not intersect
itself anywhere between its endpoints.
Example 49. A circle is a simple curve but the figure 8 is not.
Theorem 2.6.1. Green’s Theorem
Let C be a positively oriented, piecewise-smooth simple closed curve in the plane and
let D be the region bounded by C. If P and Q have continuous partial derivatives on
an open region that contains D, then∫C
Pdx + Qdy =
∫∫D
(∂Q
∂x− ∂P
∂y
)dA
Remarks
(Other formulas for Green’s Theorem) Let F = P i + Qj. Then
(a)
∫C
Pdx + Qdy =
∫ ∫D
(curl F) · k dA
(b)
∫C
−Qdx + Pdy =
∫ ∫D
(∂P
∂x+
∂Q
∂y
)dA =
∫ ∫D
div F dA
Example 50. Evaluate the line integral
∫C
(3y− esin x)dx + (7x +√
y4 + 1)dy, where
C is the circle x2 + y2 = 9.
Answer. 36π
Example 51. Verify that Green’s Theorem is true for the line integral∫C
xydx + x2dy,
where C is the triangle with vertices (0, 0), (1, 0), and (1, 2).
44
Answer. 2/3
Reading Assignment 11. Verify that Green’s Theorem is true for the line integral∫C
y2dx + x2dy,
where C is the boundary curve of the region R lying between y = x and 4x = y2.
Answer. Let I1 =
∫C
Pdx+Qdy =
∫C
y2dx+x2dy and I2 =
∫∫R
(∂Q
∂x− ∂P
∂y
)dA =∫∫
R
(2x− 2y)dA.
We have to verify that I1 = I2.
Calculate I2 =
∫∫R
(2x− 2y)dA :
R : y ≤ x ≤ y2
4, 0 ≤ y ≤ 4
I2 =
∫ 4
0
∫ y
y2/4
(2x − 2y)dxdy =
∫ 4
0
[x2 − 2yx
]y
y2/4dy =
∫ 4
0
(−y4
16+
y3
2− y2
)dy =
· · · = −32
15
Calculate I1 =
∫C
y2dx + x2dy :
C = C1 + C2
C1 : y = x from x = 0 to x = 4
On C1 : dy = dx∫C1
y2dx + x2dy =
∫ 4
0
x2dx + x2dx = · · · = 128/3
C2 : 4x = y2 from y = 4 to y = 0
On C2 : 4dx = 2ydy ⇒ dx =1
2ydy∫
C2
y2dx + x2dy =
∫ 0
4
y2
(1
2ydy
)+ (y2/4)2dy = · · · = −224/5
I1 =
∫C1
+
∫C2
= (128/3) + (−224/5) = −32
15= I2 as expected.
45
2.7 Surface Area and Surface Integrals
2.7.1 Surface Area
Let R be the region on the xy-plane. Let S be the surface above R with equation
z = f(x, y). If f has continuous first partial derivatives on R, then the area of S is
A(S) =
∫ ∫R
√fx
2 + fy2 + 1 dA.
Example 52. Find the area of the surface S if it is the part of the plane x+y+z = 1
that lies in the first octant.
Answer.√
3/2
Reading Assignment 12. Find the surface area of the portion of the paraboloid
z = 4− x2 − y2 lying above the xy-plane.
Answer. The area of the surface S is A(S) =
∫ ∫R
√1 + f 2
x + f 2y dA.
f(x, y) = 4− x2 − y2 ⇒ fx = −2x, fy = −2y ⇒ A(S) =
∫ ∫R
√1 + 4x2 + 4y2 dA
The vertical projection of S onto the xy−plane is the circular region R = {(x, y) : x2+
y2 ≤ 4}.In polar coordinates, R = {(r, θ) : 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
Using the substitution x2 + y2 = r2, dA = rdrdθ,
A(S) =
∫ 2π
0
∫ 2
0
√1 + 4r2 rdrdθ = 2π
[ 1
12(1 + 4r2)3/2
]2
0= π(173/2 − 1)/6
46
2.7.2 Surface Integrals
Let R be the region on the xy−plane. Let S be the region above R with equation z =
f(x, y). If f has continuous first partial derivatives on R and g(x, y, z) is continuous
on S, then the surface integral of g on S is defined as∫ ∫S
g(x, y, z) dS =
∫ ∫R
g(x, y, f(x, y))√
1 + fx2 + fy
2 dA.
Remarks
(a)
∫ ∫S
dS = area of S.
(b) dS =√
1 + fx2 + fy
2 dA.
Example 53. Evaluate the surface integral
∫ ∫S
z2 dS where S is the part of the
cylinder z =√
1− x2 that lies above the square with vertices (−1,−1), (1,−1), (−1, 1)
and (1, 1).
Hint : You may assume
∫ √1− x2 dx =
x
2
√1− x2 +
1
2sin−1 x + C
Answer. π
Example 54. A curved lamina is the portion of the paraboloid z = x2 + y2 below the
plane z = 1 and has density σ(x, y, z) = 6√
1 + 4z. Find the mass of the lamina.
Answer. 18π
47
Reading Assignment 13.
Let S be the portion of the cone z = 4−2√
x2 + y2 between z = 0 and z = 4. Evaluate
the surface integral
∫ ∫S
3z√
5x2 + 5y2 dS.
Answer. S : z = 4− 2√
x2 + y2, 0 ≤ z ≤ 4
zx = − 2x√x2 + y2
, zy = − 2y√x2 + y2
dS =√
1 + z2x + z2
ydA =
√1 +
4x2
x2 + y2+
4y2
x2 + y2dA
I =
∫ ∫S
3z√
x2 + y2 dS =
∫ ∫R
3(4− 2
√x2 + y2
) √5x2 + 5y2
√1 +
4x2
x2 + y2+
4y2
x2 + y2dA
=
∫ ∫R
3(4− 2
√x2 + y2
)3√
x2 + y2dA
Using polar coordinates,
R : 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
I = · · · =∫ 2π
0
∫ 2
0
9(4r2 − 2r3)drdθ = · · · = 48π
2.7.3 Surface Integrals of Vector Fields
Definition 2.7.1. Oriented Surfaces
A surface S is orientable if it is possible to choose a unit normal vector n at each
point (x, y, z) on S (except possibly at any boundary points) in such a way that n
varies continuously (have no abrupt changes in direction) as we traverse along any
curve on S. Once n has been chosen, we say that we have oriented the surface.
Note: A surface can be oriented only if it has 2 sides, the process of orientation
consists essentially in choosing which side we will call ”positive” and which ”negative”.
(If the surface is closed, then it is natural to speak of the ”inside” and ”outside”.)
Example 55. The 2 possible orientations of a sphere are by inward unit normals or
outward unit normals.
48
Example 56. A Mobius strip is a non-orientable surface since it has only 1 side.
Note : We will exclude the non-orientable surfaces from further consideration. Hence
forth, whenever we say ”surface”, we mean an ”orientable surface”.
49
Definition 2.7.2. If F is a continuous vector field defined on an oriented surface S
with unit normal vector n, then the surface integral of F over S is
∫ ∫S
F · n dS.
This integral is also called the flux of F across S.
Note 7.
(a) Recall that the gradient vector of a scalar function h(x, y, z) at a point P (x0, y0, z0)
is the vector
∇h(x0, y0, z0) = hx(x0, y0, z0)i + hy(x0, y0, z0)j + hz(x0, y0, z0)k.
(b) At any point (x0, y0, z0) in the domain of h(x, y, z), ∇h(x0, y0, z0) is orthogonal
to the surface h(x, y, z) = h(x0, y0, z0).
Example 57. Evaluate the flux integral I =
∫ ∫S
F · n dS for the vector field
F = −yi + xj + 3zk, where S is the hemisphere z =√
16− x2 − y2 with upward
orientation.
Answer. 128π
2.8 Stokes’ Theorem
Definition 2.8.1. A surface S : G(x, y, z) = C is smooth if the normal vector ∇G
is continuous and never vanishes on S. (It has no ”corners”.)
It is said to be piecewise smooth if it consists of a finite number of smooth parts
”joined together”
For example, the surface of a sphere is smooth whereas the surface of a cube is piece-
wise smooth.
Theorem 2.8.1. Stokes’ Theorem
Let S be an oriented piecewise-smooth surface that is bounded by a simple, closed,
piecewise-smooth boundary curve C with positive orientation. If F is a continuously
50
differentiable vector field on an open set that contains S, then∮C
F · dr =
∫ ∫S
curlF · n dS.
Example 58. Use Stokes’ Theorem to compute the line integral
∮C
F · dr, where
F(x, y, z) = 4yi− 3zj + xk and C is the triangle having vertices at (1, 0, 0), (0, 1, 0),
and (0, 0, 1), counterclockwise when viewed from above.
Answer. −1
Remark 5. If S1 and S2 are oriented surfaces with the same oriented boundary curve
C and both satisfy the hypotheses of Stokes’ Theorem, then∫ ∫S1
curlF · n dS =
∮C
F · dr =
∫ ∫S2
curlF · n dS.
This fact is useful when it is difficult to integrate over one surface but easy to integrate
over the other.
Example 59. Let S be the part of the paraboloid z = 5 − x2 − y2 that lies above
the plane z = 1, oriented upward. Verify Stokes’ Theorem for the vector field F =
2zi + 3xj + 5yk.
Answer. Both integrals are equal to 12π.
Reading Assignment 14.
Let F = (z − y)i + (x− z)j + (x− y)k.
Use Stokes’ Theorem to evaluate the surface integral I =
∫ ∫S
curlF · n dS, where S
is the hemisphere z =√
1− x2 − y2.
Answer. By Stokes’ Theorem,
I =
∮C
F · dr
51
where C is the circle x2 + y2 = 1, z = 0.
C : x = cos t, y = sin t, z = 0, 0 ≤ t ≤ 2π
r = xi + yj + xk ⇒ r′ = − sin ti + cos tj
F = (z − y)i + (x− z)j + (x− y)k = (0− sin t)i + (cos t− 0)j + (cos t− sin t)k
F · r′ = sin2 t + cos2 t = 1
I =
∫ 2π
0
F · r′dt =
∫ 2π
0
dt = 2π
2.9 Gauss’ Divergence Theorem
Theorem 2.9.1. Gauss’ Divergence Theorem
Let S be a sectionally smooth surface. Let V be the solid enclosed by S. Let F be
a vector field defined on S. If the component functions of F have continuous first
derivatives in V , then ∫ ∫S
F · n dS =
∫ ∫ ∫V
divF dV
where n is the outward unit normal.
Example 60. Let F = 3xyi+y2j−x2y4k, and S be the surface of the tetrahedron with
vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1). Use the divergence theorem to evaluate
the surface integral
∫ ∫S
F · n dS, where n is the outward-pointing unit normal.
Answer. 5/24
Example 61. Verify the divergence theorem for F = xi + yj + zk over the sphere
x2 + y2 + z2 = a2.
Answer. Both integrals are equal to 4πa3.
52
Reading Assignment 15.
Verify the divergence theorem for F = x2i + y2j + z2k over the unit cube bounded by
x = 1, y = 1, z = 1, and the coordinate planes.
Answer. Let I1 =
∫ ∫S
F · n dS and I2 =
∫ ∫ ∫V
divF dV.
We have to verify that I1 = I2.
(i) divF =∂
∂x
(x2
)+
∂
∂y
(y2
)+
∂
∂z
(z2
)= 2x + 2y + 2z
I2 =
∫ ∫ ∫V
(2x + 2y + 2z)dV =
∫ 1
0
∫ 1
0
∫ 1
0
(2x + 2y + 2z)dzdydx = · · · = 3
(ii) I1 =
∫S1
F · ndS + · · ·+∫
S6
F · ndS,
S = S1 + · · ·+ S6, where S1, . . . , S6 are the 6 faces of the cube.
On S1 : x = 0, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, we have
n = outward unit normal to S1 = −i
F = x2i + y2j + z2k = y2j + z2k
F · n = 0
Hence, J1 =
∫ ∫S1
F · ndS = 0
On S2 : x = 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, we have
n = outward unit normal to S2 = i
F = i + y2j + z2k
F · n = 1
Hence, J2 =
∫ ∫S2
dS = area of S2 = 1.
The remaining calculations are summarized in the following table:
S S1 : x = 0 S2 : x = 1 S3 : y = 0 S4 : y = 1 S5 : z = 0 S6 : z = 1
n −i i −j j −k k
F y2j + z2k i + y2j + z2k x2i + z2k x2i + j + z2k x2i + y2j x2i + y2j + k
F · n 0 1 0 1 0 1∫SF · ndS 0 1 0 1 0 1
∴ I1 = 0 + 1 + 0 + 1 + 0 + 1 = 3 = I2 as expected.
53
54
Chapter 3
Z Transform
3.1 Introduction
Difference equations are used to model discrete-time systems. They are of growing
importance as more and more engineering systems now contain a microprocessor
or computer. For example, many factories have a production control computer to
schedule production. Such as the Laplace transform was used to solve differential
equations, we will develop the z-transform and use it to solve difference equation.
Definition 3.1.1. A sequence is an ordered set of complex numbers
{xk}∞−∞ = {. . . , x−2, x−1, x0, x1, x2, . . .}.
xk is called the kth term of the sequence.
{xk} is called a causal sequence if xk = 0 for k < 0.
Definition 3.1.2. The z transform of a given causal sequence , {xk}∞0 , is defined
as
X(z) =∞∑
k=0
xk
zk.
whenever the sum exists and where z is a complex variable, as yet undefined.
55
Some simple examples of z transforms of common discrete signals are now given.
Example 62. Determine the z-transform of the unit impulse sequence {δk}∞0 =
{1, 0, 0, . . .}.
Answer. 1
Example 63. Determine the z-transform of the unit step sequence {uk = 1}∞0 =
{1, 1, 1, . . .}.
Answer.z
z − 1
Example 64. Determine the z-transform of the exponential sequence {xk = ak}∞0where a is a constant.
Answer.
Example 65. Determine the z-transform of the unit ramp sequence {k}
Answer. Z{k} =z
(z − 1)2
Reading Assignment 16. Determine the z-transform of the sequence {xk = kak−1}where a is a constant.
Answer. From the previous example, we see that
Z{ak} =∞∑
k=0
ak
zk=
z
z − a
Differentiating wrt a :∞∑
k=1
kak−1
zk=
∞∑k=0
kak−1
zk=
z
(z − a)2
⇒ Z{kak−1} =z
(z − a)2
Example 66 (Sampling a continuous signal). Consider the continuous signal
f(t) = e−t defined for t ≥ 0. Suppose we sample or measure the values of f(t)
at intervals of time, T , we obtain a discrete signal defined by the sequence xk =
f(kT ) = e−kT , for k = 0, 1, 2, . . . . Find the z transform of the resulting signal.
56
Answer. The transform is Z{(e−T )k} =z
z − e−T
Reading Assignment 17. The continuous-time signal f(t) = 5 sin 5t, t ≥ 0 is
sampled at equal intervals of t =π
10.
(a) Write down the general term of the sequence of samples.
(b) Find the z transform of the sequence.
Answer.
(a) The general term of the sampled sequence xk = f(k
π
10
)= 5 sin
kπ
2, k =
0, 1, 2, . . . .
(b) X(z) =5z
z2 + 1(Table)
Exercise 16. The continuous-time signal f(t) = 4t, t ≥ 0 is sampled at t =
2k, k = 0, 1, 2, . . . .
(a) Write down the general term of the sequence of samples.
(b) Find the z transform of the sequence.
Answer. xk = 8k, k = 0, 1, 2, . . . ; X(z) =8z
(z − 1)2
57
3.2 Properties of The Z Transform
The z-transform has many important and useful properties which are useful for ma-
nipulating and solving problems.
Theorem 3.2.1 (Linearity). Let xk and yk be two sequences with z transforms X(z)
and Y (z) respectively. The z-transform operator Z is a linear operator. That is ,
Z{axk + byk} = aZ{xk}+ bZ{yk} = aX(z) + bY (z)
for any constants a and b.
Example 67. Use the fact Z{ak} =z
z − aand the linear property to find the z
transform of the causal sequence
(a) cos(ak)
(b) sin(ak)
Hint : cos(ak) =1
2
(eiak + e−iak
), sin(ak) =
1
2i
(eiak − e−iak
)Answer. (a)
z(z − cos a)
z2 − 2z cos a + 1(b)
z sin a
z2 − 2z cos a + 1
Theorem 3.2.2. (The First Shift Theorem)
Z{xk−m} =1
zmZ{xk} where m is a positive integer.
Proof. Z{xk−m} =∞∑
k=0
xk−m
zk= z−m
∞∑k=m
xk−m
zk−m, (since xk−m = 0 for k−m < 0 or k <
m)
= z−m
∞∑k=0
xk
zk=
1
zmZ{xk}
Example 68. The causal sequence {xk} is generated by
xk = ak, k ≥ 0.
where a is a constant. Determine the z transform of the shift sequence {xk−m} where
m is a positive integer.
58
Answer. Z{xk−m} =1
zm
z
z − a
Theorem 3.2.3 (The Second Shift Theorem).
(a) Z{xk+1} = zZ{xk} − zx0 = zX(z)− zx0
(b) Z{xk+2} = z2X(z)− z2x0 − zx1
(c) Z{xk+a} = zaZ{xk} −a−1∑n=0
xnza−n where a is a positive integer.
Proof. (a) Z{xk+1} =∞∑
k=0
xk+1
zk= z
∞∑k=0
xk+1
zk+1= z
∞∑m=1
xm
zm= z
( ∞∑m=0
xm
zm− x0
)= z[Z{xk} − x0] = zZ{xk} − zx0 = zX(z)− zx0
(b) Z{xk+1} = zZ{xk} − zx0 ⇒ Z{yk+1} = zZ{yk} − zy0, yk = xk+1
⇒ Z{xk+2} = zZ{xk+1} − zx1 = z(zZ{xk} − zx0)− zx1 = z2X(z)− z2x0 − zx1
(c) Prove by induction.
Example 69. Use the fact Z{ak} =z
z − aand the Second Shift Theorem to find
Z{ak+2}.
Answer.a2z
z − a.
3.3 More Properties of The Z Transform
Theorem 3.3.1 (Multiplication by ak). If Z{xk} = X(z), then for a constant a,
Z{akxk} = X(a−1z) or Z{a−kxk} = X(az).
This is also referred to as the change of scale or damping rule.
Example 70. Use the fact Z{k} =z
(z − 1)2and the damping rule to show that
Z{kak} =az
(z − a)2.
59
Answer.az
(z − a)2
Theorem 3.3.2 ( Multiplication by k).
Z{kxk} = −zd
dz[X(z)].
Example 71. Show that Z{k cos kθ} =z3 cos θ − 2z2 + z cos θ
(z2 − 2z cos θ + 1)2if Z{cos kθ} =
z(z − cos θ)
z2 − 2z cos θ + 1.
Answer.z3 cos θ − 2z2 + z cos θ
(z2 − 2z cos θ + 1)2
3.4 The Inverse Z Transform
Definition 3.4.1. If Z{xk} = X(z), then xk = Z−1{X(z)} is called the inverse z
transform of X(z).
Theorem 3.4.1. The inverse z transform is also a linear operator.
Example 72. Find Z−1{ z
(z − 7)2
}.
Answer. xk = k7k−1, k ≥ 0
Example 73. Find xk = Z−1{ 3z + 4
(z + 1)(z − 2)
}.
Answer. xk =10
32k−1 − 1
3(−1)k−1, k ≥ 1; x0 = 0
Example 74. Invert the z transform Y (z) =z
z2 + a2where a is a real constant.
Answer. yk == ak−1 sin 12kπ, k ≥ 0
60
Reading Assignment 18. Find yk = Z−1{ 36z2 − 12z
(z + 1)(z − 3)2
}.
Answer. Let Y = Z{yk} =36z2 − 12z
(z + 1)(z − 3)2.
(a) By partial fractions,Y
z=
36z − 12
(z + 1)(z − 3)2=
A
z + 1+
B
z − 3+
C
(z − 3)2
= · · · = 3
z − 3+
24
(z − 3)2− 3
z + 1
⇒ Y =3z
z − 3+
24z
(z − 3)2− 3z
z + 1.
(b) By using the Table,
yk = 3(3)k + 8k(3)k − 3(−1)k, k ≥ 0.
3.5 Solving Difference Equations
Example 75. Solve the first order difference equation
yk+1 − 5yk = 0, y0 = 7.
Answer. yk = 7(5)k, k ≥ 0
Example 76. Solve the second order difference equation
yk+2 − 5yk+1 + 6yk = 0, y0 = 7, y1 = 16.
Answer. yk = 5(2)k + 2(3)k, k ≥ 0
61
Reading Assignment 19. Solve yk+2 + 6yk+1 + 9yk = 2k with y0 = y1 = 0.
Answer. [z2Y − z2y0 − zy1] + 6[zY − zy0] + 9Y =z
z − 2
⇒ z2Y + 6zY + 9Y =z
z − 2
⇒ Y =z
(z − 2)(z + 3)2
⇒ Y
z=
1
(z − 2)(z + 3)2= · · · = 1
25
1
z − 2− 1
25
1
z + 3− 1
5
1
(z + 3)2
⇒ Y =1
25
z
z − 2− 1
25
z
z + 3− 1
5
z
(z + 3)2
⇒ yk =1
25(2)k − 1
25(−3)k +
1
15k(−3)k, k ≥ 0
Reading Assignment 20. Solve the difference equation
yk+2 − 2yk+1 + yk = 2k
given that y0 = 2 and y1 = 1.
Answer. [z2Y − z2y0 − zy1]− 2[zY − zy0] + Y =z
z − 2
[z2Y − 2z2 − z]− 2[zY − 2z] + Y =z
z − 2
(z2 − 2z + 1)Y = 2z2 − 3z +z
z − 2
Y
z=
2z − 3
(z − 1)2+
1
(z − 2)(z − 1)2
= · · · =(
2
z − 1− 1
(z − 1)2
)+
(1
z − 2− 1
z − 1− 1
(z − 1)2
)=
1
z − 2+
1
z − 1− 2
(z − 1)2
Y =z
z − 2+
z
z − 1− 2z
(z − 1)2
yk = 2k + 1− 2k, k ≥ 0
62
3.6 A Table Of Z Transform
The z transforms of some common functions are listed in the following table where a
and b are constants.
xk (k ≥ 0) Z{xk} = X(z)
δk =
{1 , k = 0
0 , k > 01
uk = 1z
z − 1ak z
z − ak
z
(z − 1)2
kak az
(z − a)2
cos akz(z − cos a)
z2 − 2z cos a + 1
sin akz sin a
z2 − 2z cos a + 1
xk−a1
zaX(z)
xk+1 zX(z)− zx0
xk+2 z2X(z)− z2x0 − zx1
kxk −zdX
dzakxk X(z/a)
Exercise 17. Use the given table to find the z transforms of
(a) cos kπ (b) (−1)kek (c) kek
Answer. (a)z
z + 1(b)
z
z + e(c)
ez
(z − e)2
Exercise 18. Find the inverse z transform of
(a)6z
3z + 1+
5z
z − 1(b)
z
z2 − z + 1(c)
z
(ez + 1)2(d)
z2
5z2 − 4z − 1
Answer. (a) 2(−1/3)k + 5 (b)2√3
sinkπ
3
(c) −k
e
(−1
e
)k
where k = 0, 1, 2, . . . (d)1
6+
(−1/5)k
30
63
64
Chapter 4
Complex Analysis
4.1 Revision
Definition 4.1.1.
(a) A complex number can be written in the Cartesian form as
a + bi,
where a, b ∈ R and i =√−1, i2 = −1.
Note that i2 = −1, i3 = −i, i4 = 1, i5 = i4i = i and so on.
(b) If z = a + bi, then the real numbers a and b are respectively called the real part
and imaginary part of z. We denote them as a = Re z and b = Im z.
(c) The number 0 + bi, b 6= 0 is called a pure imaginary number.
(d) Graphical Representation of Complex Numbers
A complex number a + bi can also be considered as an ordered pair (a, b). So we
can represent it by a point in an xy-plane called the complex plane or Argand
diagram. We call the x-axis as the real axis and the y-axis as the imaginary
axis.
65
(e) The number z̄ = a− bi is called the complex conjugate of z = a + bi. It is the
reflection of the point z about the real axis.
(f) Let z = a + bi where a, b ∈ R.
(i) The number |z| =√
a2 + b2 is called the magnitude or modulus of z.
Note : zz̄ = (a + bi)(a− bi) = a2 + b2 = |z|2
(ii) The argument of z, written arg z, is the angle θ = tan−1 b
a, a 6= 0, which
is multi-valued.
(iii) Define the principal argument or principal value of the argument as
Arg z where
−π < Arg(z) ≤ π.
Example 77. Find Arg(z) if (a) z = i, (b) z = −1, (c) z = 1 + i
Answer. Arg(i) =π
2,Arg(−1) = π,Arg(1 + i) =
π
4.
66
Theorem 4.1.1. A complex number z = x + yi can be written in polar form as
z = r(cos θ + i sin θ) = rcis θ where r = |z| =√
x2 + y2 and θ = arg(z).
Note 8.
(a) By considering the power series of the exponential function, we can show that
eiθ = cos θ + i sin θ.
This is called Euler’s formula. Thus we can now write in an exponential
form
z = reiθ.
(b) z̄ = e−iθ
Theorem 4.1.2 (De Moivre Formula).
(cos θ + i sin θ)n = cos nθ + i sin nθ, n = 0, 1, 2, 3, . . . .
Note : This formula also holds for any real number n. For example,√
cos θ + i sin θ =
cosθ
2+ i sin
θ
2.
Definition 4.1.2 (Roots of Complex Numbers).
(a) A number w is called an nth root of a complex number z if wn = z, and we write
w = z1/n.
(b) From De Moivre Formula, we can show that if n ∈ N, then
z1/n = [r(cos θ+i sin θ)]1/n = r1/n[cos
θ + 2kπ
n+i sin
θ + 2kπ
n
], k = 0, 1, 2, . . . , n−1.
Notice that there are n different values for z1/n, i.e. n different nth roots of z
provided z 6= 0.
(c) The solutions of the equation zn = 1 where n ∈ N are called the nth roots of
unity and are given by
z = cos2kπ
n+ i sin
2kπ
n= e
2kπn
i, k = 0, 1, 2, . . . , n− 1.
67
(d) If we let ω = cos2π
n+ i sin
2π
n= e
2πn
i, then the nth roots are
1, ω, ω2, . . . , ωn−1.
Geometrically, they represent the n vertices of a regular polygon of n sides in-
scribed in the unit circle x2 + y2 = 1.
Reading Assignment 21. Find the cube root of 1, i.e. solve ω3 = 1.
Answer. ω3 = 1 = cis0 ⇒ ω = 11/3cis0 + 2kπ
3, k = 0, 1, 2 ⇒ ω = cis(0), cis
(2π
3
), cis
(4π
3
)∴ ω = 1,−1
2±√
3
2i.
Reading Assignment 22. Find the complex numbers z which satisfy z2 = 4i.
Answer. z2 = 4i = 4cis(π
2+ 2kπ
)⇒ z = 41/2cis
π2
+ 2kπ
2, k = 0, 1
⇒ z = 2(
cosπ
4+ i sin
π
4
), 2
(cos
5π
4+ i sin
5π
4
)⇒ z = ±
√2(1 + i).
4.2 Loci and Regions of the Complex Plane
Definition 4.2.1.
(a) A circle with radius r and center z0 can be represented by the equation
|z − z0| = r.
It can also be represented in the parametric form
z = z0 + reit, 0 ≤ t < 2π.
(b) The set of points {z ∈ C : |z − z0| < r} is called an open disc of radius r
centered at z0.
68
(c) The set of points {z ∈ C : |z − z0| ≤ r} is called an closed disc of radius r
centered at z0.
(d) Any open disc about z0 is called a neighborhood of z0.
(e) z0 is called an interior point of a set S if there exists a neighborhood of z0
which contains only points of S.
(f) z0 is called a boundary point of a set S if every neighborhood of z0 contains
both points in S and points not in S.
(g) A set S is open if every point of S has a neighborhood containing entirely of
points in S.
(h) A set S is closed if its complement (all points not in S) is open.
(i) An open set S is connected if any 2 points of S can be joined by a polygonal
line, consisting of a finite number of line segments joined end to end, that lies
entirely in S.
(j) An open set that is connected is called a domain.
(k) A set is bounded if it can be enclosed by a circle of finite radius. Otherwise, it
is unbounded.
Example 78.
(a) The interior of a circle is open.
(b) The set of all points outside the open disc of radius r about z0 , i.e. {z : |z−z0| ≥r} is a closed set.
(c) The set {z : 0 < |z| ≤ 1} is neither open nor closed.
(d) The open set |z| < 1 is connected . The annulus 1 < |z| < 2 is also connected.
4.3 Functions of a complex variable
Definition 4.3.1. Let S be a set of complex numbers. A function f : S → Cdefined on S is a rule that assigns to each z ∈ S a complex number f(z).
69
Note : We can write
f(z) = f(x + iy) = u(x, y) + iv(x, y)
where u(x, y) = Ref(z), v(x, y) = Imf(z) are real-valued functions of the real vari-
ables x and y.
Example 79. Find Re(f) and Im(f) if f(z) = z2.
Answer. Re(f) = x2 − y2, Im(f) = 2xy
Definition 4.3.2 (Polynomial and Rational Functions).
(a) Let n be a nonnegative integer. The function
P (z) = anzn + an−1z
n−1 + · · ·+ a1z + a0,
where a0, . . . , an ∈ C, is called a polynomial function.
(b) R(z) = P (z)/Q(z) where P and Q are polynomials is a rational function.
Definition 4.3.3 (Limit). We say that the limit of f(z) as z approaches z0 exists
and is equal to a number L, written,
limz→z0
f(z) = L,
if for every ε > 0 there exists δ > 0 such that
|f(z)− L| < ε whenever 0 < |z − z0| < δ.
Remark 6. z may approach z0 from any direction.
Definition 4.3.4 (Continuity). A function f(z) is said to be continuous at a point
z0 if f(z0) exists, limz→z0
f(z) exists and
limz→z0
f(z) = f(z0).
70
Definition 4.3.5 (Differentiability). f is differentiable at z0 if the limit
f ′(z0) = limz→z0
f(z)− f(z0)
z − z0
or limδz→0
f(z0 + δz)− f(z0)
δz
exits.
Note 9. The differentiation rules of real calculus are also valid in complex differen-
tiation.
Exercise 19. Use the definition to show that f ′(z) = 2z if f(z) = z2.
Example 80. Show that f(z) = z̄ = x− iy is not differentiable everywhere.
Definition 4.3.6. [Analyticity]
(a) A function f is analytic at a point z0 if f is defined and has a derivative at
every point in a neighborhood of z0.
(b) f is analytic in a domain D if it is analytic at every point of D.
Note 10. A point at which f is not analytic is called a singular point or singu-
larity.
Example 81. f(z) =1
z − 4has a singularity at z = 4.
Example 82. f(z) =1
z2 + 4has singularities at z = ±2i.
Example 83. The polynomial functions are analytic everywhere. That is, they are
what we called the entire functions.
Example 84. f(z) = |z|2 is differentiable only at z = 0, so it is nowhere analytic.
Theorem 4.3.1. (Properties of analytic functions)
If f and g are analytic in a domain D, then f ± g , fg , and f ◦ g are analytic in D.
Similarly, f/g is analytic in D provided g(z) 6= 0 for all z ∈ D.
In particular, the rational functionp(z)
q(z)(p and q are polynomials) is analytic in any
domain throughout which q(z) 6= 0.
71
Example 85. The rational function3z + 5
z2 + 4is analytic everywhere except at the points
where z2 + 4 = 0 i.e., at the points z = ±2i.
Theorem 4.3.2. [Cauchy-Riemann Equations]
Let f(z) = u(x, y) + iv(x, y) be defined and continuous in some neighborhood of a
point z = x + iy. If f is differentiable at z, then at this point, the first-order partial
derivatives of u and v exist and satisfy the Cauchy-Riemann equations
∂u
∂x=
∂v
∂yand
∂u
∂y= −∂v
∂x. (4.1)
Corollary 4.3.3. If f is analytic in a domain D, then the first-order partial deriv-
atives of u and v exist and satisfy the Cauchy-Riemann equations at all points of
D.
Corollary 4.3.4. In this case, we can calculate using
f ′(z) = ux + ivx or f ′(z) = −iuy + vy.
Theorem 4.3.5. If the real-valued functions u(x, y) and v(x, y) have continuous first
partial derivatives that satisfy the Cauchy-Riemann equations in some domain D,
then f(z) = u(x, y) + iv(x, y) is analytic in D.
Definition 4.3.7. [Harmonic Functions]
A function h(x, y) is said to be a harmonic function if satisfies the Laplace’s equa-
tion
∇2h = hxx + hyy = 0.
Theorem 4.3.6. If f(z) = u(x, y) + iv(x, y) is analytic in a domain D, then u and
v are harmonic functions.
Note 11. The functions u(x, y) and v(x, y) are sometimes called conjugate har-
monic functions. Given one we can find the other (within an arbitrary additive
constant) so that u + iv = f(z) is analytic .
72
Example 86 (Finding a conjugate harmonic function). Verify that v(x, y) = 2xy is
harmonic and obtain the most general conjugate harmonic function u(x, y) of v.
Answer. u(x, y) = x2 − y2 + C where C is an arbitrary constant
Example 87 (Reading Assignment).
(a) Show that u(x, y) = sin x cosh y is harmonic.
(b) Find a harmonic conjugate v(x, y) for u.
Answer. (a) We need to show that uxx + uyy = 0.
u(x, y) = sin x cosh y ⇒
{ux = cos x cosh y , uxx = − sin x cosh y
uy = sin x sinh y , uyy = sin x cosh y
⇒ uxx + uyy = − sin x cosh y + sin x cosh y = 0
(b) To find a harmonic conjugate v(x, y), we solve the C-R equations:
vy = ux = cos x cosh y (4.2)
vx = −uy = − sin x sinh y (4.3)
(2)⇒ v = cos x sinh y + φ(x), (where φ(x) is a function of x)
⇒ vx = − sin x sinh y + φ′(x)
⇒ φ′(x) = 0 (by comparing with (3)) ⇒ φ(x) = C, a constant
⇒ v = cos x sinh y + C
4.4 More Elementary Functions
Definition 4.4.1.
(a) An exponential function is defined by ez = ex+iy = ex(cos y + i sin y) = exeiy.
Note : We want ez to be consistent in the case z = x ∈ R and insist that it
satisfies the properties (P1)d
dz(ez) = ez and (P2) ez1+z2 = ez1ez2 .
Example 88. Use the Cauchy-Riemann equations to show that f(z) = ez is entire,
i.e. analytic everywhere. Hence, find f ′(z).
73
Answer. Show that it satisfies the C-R equations. Then f ′(z) = ux + ivx = · · · = ez.
Definition 4.4.2 (Trigonometric Functions). We define the trigonometric functions
in terms of exponential functions as follows:
(a) sin z =eiz − e−iz
2i
(b) cos z =eiz + e−iz
2
(c) tan z =sinz
cos z
(d) cot z =cos z
sin z
We will state the following theorems without proofs.
Theorem 4.4.1 (Properties of Trigonometric Functions).
(a) ez and e−z are entire functions ⇒ cos z and sin z are entire functions,
(b) We also haved
dz(sin z) = cos z,
d
dz(cos z) = − sin z,
d
dz(tan z) = sec2 z and so
on.
(c) sin(z1 ± z2) = sin z1 cos z2 ± cos z1 sin z2
(d) cos(z1 ± z2) = cos z1 cos z2 ∓ sin z1 sin z2
(e) cos 2z = cos2 z − sin2 z, sin 2z = 2 sin z cos z
Note 12. (a) From the definition, sin z is odd whereas cos z is even.
(b) Complex trigonometric functions have all the standard properties of real trigono-
metric functions.
Definition 4.4.3 (Hyperbolic Functions). (a) cosh z =1
2(ez + e−z)
(b) sinh z =1
2(ez − e−z)
(c) tanh z =sinh z
cosh z
(d) sechz =1
cosh z
74
Note 13. All the standard formulas apply, for example
d
dz(cosh z) = sinh z,
d
dz(sinh z) = cosh z, cosh2 z − sinh2 z = 1.
Definition 4.4.4 (Complex Logarithmic Functions). w = ln z ⇔ z = ew(z 6=0 as ew 6= 0).
Thus the natural logarithmic function is the inverse of the exponential function and
can be defined by
ln z = ln |z|+ iargz = ln |z|+ i(Argz + 2nπ), n = 0,±1,±2, . . . .
Note that ln z is a multi-valued. However, the principal value of ln z defined as
Lnz = ln |z|+ iArgz, − π < Arg ≤ π
is a well defined (single-valued) function.
Note 14. If z = x > 0 (real and positive), then Argz = 0 and ln z = ln x, a real
logarithmic function.
Example 89. Find all the values for (a) ln 2i (b) ln(−3). What are their principal
values?
Answer. (a) ln 2i = ln 2 + i(π
2± 2nπ
); Ln(2i) = ln 2 + i
π
2(b) ln(−3) = ln 3 + i
(π±
2nπ), n = 0, 1, 2, . . . ; Ln(−3) = ln 3 + iπ
Example 90 (Reading Assignment). Find all the values for ln(2 − 2i). What is its
principal value?
Answer. (a) Let w = ln z where z = 2− 2i. Then
(i) |z| = · · · =√
8 and
(ii) arg(z) = · · · = −π
4± 2kπ, k = 0, 1, 2, . . . .
(b) Hence all the values of w are
ln(2− 2i) = ln z = ln |z|+ iarg(z)
= ln√
8 + i(− π
4± 2kπ
), k = 0, 1, 2, . . . .
75
(c) Its principal value is Lnz = ln√
8− iπ
4(recall that −π < Arg(z) ≤ π.)
Definition 4.4.5. Any z for which f(z) = 0 is called a solution or a root of that
equation or a zero of f.
Example 91. The roots of the quadratic equation az2 + bz + c = 0 are given by
z =−b±
√b2 − 4ac
2a.
Example 92. Solve for z : ln z = −iπ2
Answer. −i
Exercise 20. Solve z2 − 2(i− 2)z − 4(i + 2) = 0.
Answer. (i− 2)±√
11
Exercise 21. Find the zeros of f(z) = sin z.
Answer. z = nπ, n = 0,±1,±2, . . . .
Definition 4.4.6 (General Powers of Complex Numbers). Define zc = ec ln x where
z 6= 0, c are any complex numbers.
This is in general multi-valued (since ln z is).
We define its principal value as ecLnz.
Example 93. Find all the values for ii.
Answer. e−(π
2± 2nπ
), n = 0, 1, 2, . . .
Example 94 (Reading Assignment). Find all the values for (−1 + i√
3)1/2 .
Answer. (a) By definition, w = (−1 + i√
3)1/2 = exp[1
2ln(−1 + i
√3]
(b) ln(−1 + i√
3) = · · · = ln 2 + i(
2π3± 2kπ
), k = 0, 1, 2, . . .
76
(c) w = exp[1
2ln 2 +
(π
3± kπ
)] = eln
√2e
i(π3±kπ
)=√
2[cos
(π
3± kπ
)+ i sin
(π
3± kπ
)]=√
2[cos
π
3cos kπ+i sin
π
3cos kπ
]using the identities cos(A+B) = · · · , sin(A+
b) = · · ·= ±
√2[cos
π
3+ i sin
π
3
], cos(kπ) = (−1)k = ±1
= ±√
2[1
2+ i
√3
2
]= ±[
√2
2+ i
√6
2]
77
4.5 Complex Integration
Definition 4.5.1.
(a) A curve C in the complex plane may be described by a parametric representation
z(t) = x(t) + iy(t), a ≤ t ≤ b.
(b) C is smooth if it has a continuous and nonzero derivativedz
dtat each point.
(c) C is piecewise smooth if it consists of finitely many smooth curves joined end
to end.
Definition 4.5.2. Let f(z) be defined on a piecewise smooth curve
C : z(t) = x(t) + iy(t), a ≤ t ≤ b.
Partition C into n parts by the points
z0 = a, z1, . . . , zn−1, zn = b
such that the greatest |4zk| = |zk − zk−1| → 0 as n →∞. Then the line integral of
f(z) along the curve C, denoted
∫C
f(z)dz, is defined as
limn→∞
n∑k=1
f(zk)4zk
provided the limit exists.
Theorem 4.5.1 (Existence Theorem). The line integral exists if C is piecewise con-
tinuous and f(z) is continuous on C.
Note 15. (Connection between real and complex line integrals)
If f(z) = u(x, y) + iv(x, y), then∫C
f(z)dz =
∫C
udx− vdy + i[ ∫
C
udy + vdx]
where both of the integrals on the right are real line integrals.
78
Example 95.
Evaluate the line integral I =
∫C
z̄ dz along the line segment C from 0 to 2i.
Answer. 2
Reading Assignment 23.
Evaluate the line integral
∫C
z2dz along the line segment C from 1 + 2i to 2 + 4i.
Answer. C : y = 2x, 1 ≤ x ≤ 2 ⇒ dy = 2dx∫C
z2dz =
∫C
(x+iy)2(dx+idy) =
∫C
[(x2−y2)dx−2xydy]+i
∫C
[2xydx+(x2−y2)dy]
=
∫ 2
1
[(x2 − (2x)2)dx− 2x(2x)2dx] + i
∫ 2
1
[2x(2x)dx + (x2 − (2x)2)(2dx)]
=
∫ 2
1
(−11x2)dx− i
∫ 2
1
(2x2)dx = −77
3− i
14
3
Theorem 4.5.2 (Some Properties of
∫C
f(z)dz).
(a) If C = C1 + C2, then
∫C
f(z)dz =
∫C1
f(z)dz +
∫C2
f(z)dz.
(b) Reverse the sense of integration, reverses the sign of the integral:∫−C
f(z)dz = −∫
C
f(z)dz
(c)
∫C
(k1f1(z) + k2f2(z)
)dz = k1
∫C
f1(z)dz + k2
∫C
f2(z)dz where k1 and k2 are
constants.
79
4.6 Two Integration Methods
Theorem 4.6.1 (First Method : Indefinite Integration of Analytic Func-
tions).
Let f be analytic in a simply-connected domain D. Then there is an indefinite integral
F of f in D, i.e. an analytic function F such that F ′(z) = f(z) in D, and for every
path C in D joining two points z0 and z1 in D, we have∫C
f(z)dz =
∫ z1
z0
f(z)dz =[F (z)
]z1
z0
= F (z1)− F (z0).
Note 16.
(a) A connected region D in the xy-plane is simply connected if every simple closed
curve in D encloses only points in D. Roughly speaking, a connected region D
simply connected if it has no holes.
The interior of a circle is simply connected whereas the annulus 1 < |z| < 2 is
not.
(b) This method is restricted to analytic functions.
(c) The value of the integral
∫C
f(z)dz depends only on the initial point z0 and the
terminal point z1 but not on the choice of the path of integration C in D. We
say, the integral is independent of path in D.
Example 96. (Integral of an analytic function)
Evaluate I =
∫C
2z dz if C is any path from 0 to 2i.
Theorem 4.6.2. (Second Method : Integration Along A path)
Let C : z(t) = x(t) + iy(t) be a piecewise smooth path parametrized by t with
a ≤ t ≤ b. Then
∫C
f(z)dz =
∫ b
a
f(z(t))z′(t)dt.
Note 17. This method is not restricted to analytic functions, but applies to any
continuous complex function.
80
Example 97. (Integral of a non-analytic function)
Evaluate I =
∫C
z̄dz if C is the counterclockwise unit circle.
Comment : As f(z) = z̄ is not analytic, we cannot use the first method to evaluate
this integral.
Answer. 2πi
Example 98. Let f(z) = (z− z0)m where m is an integer. Evaluate the line integral
I =
∫C
f(z)dz where and C is the (anticlockwise) circle of center z0 and radius ρ.
Answer. I = 2πi, if m = −1; 0 otherwise.
Theorem 4.6.3 (Cauchy’s Integral Theorem ). If f(z) is analytic in a simply con-
nected domain D, then for every simple closed path C in D,
∮C
f(z)dz = 0.
Example 99. (No singularities)
Evaluate I =
∫C
ezdz for any closed path C.
Example 100 (Singularities outside contour). Evaluate
∫C
dz
z2 − 4where C is the
unit circle.
Theorem 4.6.4 (Some Consequences of Cauchy’s Integral Theorem).
(a) Let f(z) be analytic in a simply connected domain D. Let C be a path inD with
initial and final points, say A and B respectively. If C ′ is another path in D with
the same initial and final points, then∫C
f(z)dz =
∫C′
f(z)dz.
That is, the integral
∫C
f(z)dz is independent of path.
(b) The Principle Of Deformation Of Path
If C1 is obtained from C by continuous deformation of path without passing
through any point where f(z) is not analytic, then
∫C
f(z)dz =
∫C1
f(z)dz.
81
(c) The above principle can be extended to the case where f(z) has a finite number
of singularities, say, z1, . . . , zn inside C. By introducing n circles C1, . . . , Cn to
enclosed each singularity, we can show that∮C
f(z)dz =
∮C1
f(z)dz + · · ·+∮
Cn
f(z)dz.
Example 101. Evaluate I =
∮C
1
zdz around any simple closed curve
(a) containing the origin;
(b) not containing the origin.
Answer. (a)2πi (b) 0
82
Theorem 4.6.5. [Cauchy’s Integral Formula]
If f(z) is analytic in a simply connected domain D, then for any point z0 ∈ D and
any simple closed path C in D which encloses z0∮C
f(z)
z − z0
dz = 2πif(z0)
where C is taken anticlockwise.
By differentiating the above equation repeatedly n times, we can extend Cauchy’s
Integral Formula to ∮C
f(z)
(z − z0)n+1dz =
2πi
n!f (n)(z0).
Note 18. This formula provides a method for evaluating the integral around a closed
path of a function g(z) which has a single simple pole at z = z0, by writing g(z) =f(z)
z − z0
where f is analytic within and on C.
Example 102. Evaluate
∮C
sin z
z − π2
dz if C : |z| = 3 , oriented counterclockwise.
Answer. = 2πi.
Example 103.
Let a be a nonzero constant. Find
∫C
2z
z2 − a2dz where C is any simple closed curve
(a) enclosing both the points a and −a;
(b) enclosing a but excluding −a.
Answer. (a) 4πi (b) 2πi
83
Theorem 4.6.6 (Taylor’s Theorem). If f(z) is analytic in a domain D and z0 is any
point in D, then there exists a unique power series (the Taylor series) representing
f(z), i.e.
f(z) =∞∑
n=0
an(z − z0)n,
where an = f (n)(z0)/n! and this representation is valid in (at least) the largest open
disc of center z0 lying in D.
In general, the theorem is valid up to the nearest singularity of f(z).
Example 104. The Maclaurin expansion of ez is
ez =∞∑
n=0
zn
n!= 1 + z +
z2
2!+
z3
3!+ · · · .
Example 105. The Maclaurin expansion of1
1− zis the geometric series
1
1− z=
∞∑n=0
zn = 1 + z + z2 + · · · .
It is valid for |z| < 1.
Theorem 4.6.7 (Laurent Series Theorem). If f(z) has an isolated singularity at
z = z0, then it has a Laurent series
f(z) =∞∑
n=0
an(z − z0)n +
b1
(z − z0)+
b2
(z − z0)2+ · · · ,
valid for 0 < |z − z0| < d where d is the distance to the nearest singularity.
Note 19. A singularity z0 of f is called an isolated singularity if there exists a
neighborhood of z0 which does not contain any other singularity of f.
Example 106. f(z) = tan z has isolated singularities at z = ±π
2,±3π
2, . . . .
Example 107. The Laurent expansion of f(z) =ez
z3about z = 0 is
f(z) =∞∑
n=0
zn/n!
z3=
1
z3+
1
z2+
1
2!
1
z+
1
3!+
z
4!+
z2
5!+ · · · .
84
Corollary 4.6.8 (Corollary to Laurent Series Theorem). If the contour C lies in
the region of validity of the Laurent series for f(z) about an isolated singularity z0,
namely
f(z) =∞∑
n=0
an(z − z0)n +
b1
(z − z0)+
b2
(z − z0)2+ · · · ,
then
∮C
f(z)dz = 2πi · b1 = 2πi·residue of f(z) at z0.
Thus, calculation of residues is equivalent to evaluation of integrals.
Theorem 4.6.9. (Calculation of Residues)
(a) Simple pole at z0
b1 = limz→z0
(z − z0)f(z)
(b) A pole of order m ≥ 2 at z = z0
b1 =1
(m− 1)!limz→z0
dm−1
dzm−1
{(z − z0)
mf(z)}
. (note : OK for m = 1 too.)
Note 20.
If f(z) has a pole of order m at z = z0, then its LS about z0 is
f(z) =bm
(z − z0)m+
bm−1
(z − z0)m−1+ · · ·+ b1
(z − z0)1+
∞∑n=0
an(z − z0)n (4.4)
Example 108. f(z) =9z + i
z(z2 + 1)has a simple pole at z = i. Find the residue of f(z)
at z = i.
Answer. −5i
Example 109. Determine the residues of f(z) =z
(z − 1)(z + 1)2at each pole.
Answer. Res(1) = 1/4, Res(−1) = −1/4
Theorem 4.6.10 (Residue Theorem). Let f(z) be analytic on and inside a simple
closed curve C except for a finite number of isolated singularities of f(z) at points
z1, z2 , . . . , zk inside an anticlockwise C. Then∮C
f(z)dz = 2πi
k∑j=1
[Residuef(z) at z = zj
]= 2π
[i(sum of residues inside C)
].
85
Example 110.
Evaluate the integral
∮C
4− 3z
z2 − zdz counterclockwise around any simple closed path C
such that
(a) 0 and 1 are inside C
(b) 0 is inside, 1 is outside
Answer. (a) 2πi(−4 + 1) = −6πi (b) −8πi
Reading Assignment 24.
Determine the residues of f(z) =1
z3(z2 + 2z + 2)at each pole.
Answer.
Find the poles of f(z) :
z3(z2 + 2z + 2) = 0 ⇒ z = 0, z =−2±
√22 − 4(1)(2)
2= −1± i.
That is, f has a pole of order 3 at z = 0, and two simple poles at z = −1± i.
Res[f(z), z = 0] =1
2!limz→0
d2
dz2
{z3f(z)
}=
1
2!limz→0
d2
dz2
{ 1
z2 + 2z + 2
}= · · ·
= limz→0
−(z2 + 2z + 2)2 + 2(z + 1)(z2 + 2z + 2)(2z + 2)
(z2 + 2z + 2)4=
1
4.
Res[f(z), z = −1 + i] = limz→−1+i
{[z − (−1 + i)]f(z)
}= lim
z→−1+i
1
z3(z + 1 + i)=
1
(−1 + i)3(2i)=
1
(2i + 2)(2i)=
1
−4 + 4i=−1− i
8
since (i− 1)3 = i3 + 3i2(−1) + 3i(−1)2 − 1 = −i + 3 + 3i− 1 = 2i + 2.
Res[f(z), z = −1 − i] = limz→−1−i
{[z − (−1 − i)]f(z)
}= lim
z→−1−i
1
z3(z + 1− i)=
1
(−1− i)3(−2i)=
1
(2− 2i)(−2i)=
1
−4− 4i=−1 + i
8
since (−i− 1)3 = (−i)3 + 3(−i)2(−1) + 3(−i)(−1)2 − 1 = i + 3− 3i− 1 = 2− 2i
86
Reading Assignment 25.
Evaluate the integral I =
∮C
1
(z − 1)(z + 2)2dz if C is the counterclockwise circle
(a) |z| = 0.5
(b) |z| = 1.5
(c) |z| = 2.5
Answer.
f(z) =1
(z − 1)(z + 2)2has a simple pole at z = 1 and a double pole at z = −2.
The residue at z = 1 is r1 = · · · = 1/9
The residue at z = −2 is r2 = · · · = −1/9
(a) Both of the poles are outside C : |z| = 0.5, so f is analytic inside C. Hence I = 0.
(b) Only the pole z = 1 is inside C : |z| = 1.5, so
I = 2πi(the residues of the poles inside C) = 2πir1 =2πi
9(c) Both the pole are inside C : |z| = 2.5, so
I = 2πi(the residues of the poles inside C) = 2πi(r1 + r2) = 0
Exercise 22.
Show that ∮Γ
sin(3z)
(z + i)4dz = −9πi cosh(3)
if Γ is the anticlockwise triangle having vertices 0, −1− 2i and 1− 2i.
Hint : cos(z) = sin(x + iy) = cos(x) cosh(y)− i sin(x) sinh(y).
Exercise 23.
Evaluate the integral I =
∮C
Re(z
i
)dz if C is the counterclockwise circle |z| = 2.
Answer. −4π
Exercise 24.
Evaluate the integral I =
∫C
6z2 cos(z3)dz if C is any smooth curve from 0 to i.
Answer. −2i sinh(1)
87
88
Chapter 5
The Fourier Integral and Fourier
Transforms
5.1 Introduction
Recall that we can represent a periodic function in terms of the sines and cosines
by means of a Fourier series. We cannot do that to a non-periodic function (eg.
the noise and pulse signals). But, we can still represent g in terms of the sines and
cosines using an integral (instead of a summation!) called the Fourier integral. The
complex form of Fourier integral will then give us the Fourier transform. Both the
Fourier integral and Fourier transform are majors tools use to solve boundary and
initial value problems in engineering and other fields.
89
Definition 5.1.1. The Fourier integral of a function f defined on (−∞,∞) is
given by ∫ ∞
0
[A(ω) cos ωx + B(ω) sin ωx]dω,
where
A(ω) =1
π
∫ ∞
−∞f(x) cos ωxdx
B(ω) =1
π
∫ ∞
−∞f(x) sin ωxdx.
Example 111. Find the Fourier integral of
f(x) =
0 , x < 0
1 , 0 < x < 2
0 , x > 2
Answer. f ∗(x) =
∫ ∞
0
1
ωπ
[sin 2ω cos ωx + (1− cos 2ω) sin ωx
]dω
Reading Assignment 26. Find the Fourier integral of
f(x) =
0 , x < 0
1 , x = 0
2e−x , x > 0
Answer. Using the results,∫eax cos bxdx =
eax
a2 + b2[a cos bx + b sin bx] + C
∫eax sin bxdx =
eax
a2 + b2[a sin bx− b cos bx] + C
A(ω) =1
π
∫ ∞
−∞f(x) cos ωxdx = · · · = 1
π
∫ ∞
0
2e−x cos ωxdx
=2
π
[ e−x
1 + ω2(− cos ωx + ω sin ωx)
]∞0
=2
π
[0− 1
1 + ω2(−1)
]=
2
π
1
1 + ω2
90
B(ω) =1
π
∫ ∞
−∞f(x) sin ωxdx = · · · = 1
π
∫ ∞
0
2e−x sin ωxdx
=2
π
[ e−x
1 + ω2(− sin ωx− ω cos ωx)
]∞0
=2
π
[0− 1
1 + ω2(0− ω)
]=
2
π
ω
1 + ω2.
The Fourier integral f ∗(x) = · · · = 2
π
∫ ∞
0
cos ωx + ω sin ωx
1 + ω2dω.
Definition 5.1.2.
Even and Odd Functions
(a) A function f is called even if f(−x) = f(x) ∀ x ∈ D(f).
(i) The functions 1, x2, x4, . . . are even.
(ii) cos x is even.
(b) A function f is called odd if f(−x) = −f(x) ∀ x ∈ D(f).
(i) The functions x, x3, x5, . . . are odd.
(ii) sin x is odd.
Remark 7.
It follows from the definitions that
(a) the product of two even functions is even
(b) the product of two odd functions is even
(c) the product of an even and an odd function is odd
Theorem 5.1.1. Integral Properties of Even and Odd Functions
(a) If f is an even integrable function defined on [−L, L], then∫ L
−L
f(x) dx = 2
∫ L
0
f(x) dx.
(b) If f is an odd integrable function defined on [−L, L], then∫ L
−L
f(x) dx = 0.
91
Definition 5.1.3.
(a) If f is an even function, then its Fourier integral is the Fourier cosine integral∫ ∞
0
A(ω) cos ωxdω,
where A(ω) =2
π
∫ ∞
0
f(x) cos ωxdx.
(b) If f is an odd function, then its Fourier integral is the Fourier sine integral∫ ∞
0
B(ω) sin ωxdω,
where B(ω) =2
π
∫ ∞
0
f(x) sin ωxdx.
Example 112. Find the Fourier integral of f(x) =
{1 , |x| ≤ 1
0 , |x| > 1
Answer. f ∗(x) =2
π
∫ ∞
0
cos ωx sin ω
ωdω.
92
5.2 Fourier Transform
Definition 5.2.1.
(a) Suppose
∫ ∞
−∞|f(t)|dt converges. Then the Fourier transform of f(t) is defined to
be the function
F[f(t)](ω) = F (ω) =
∫ ∞
−∞f(t)e−iωtdt.
(b) Sometimes we will also write F (ω) as f̂(ω).
(c) The variable ω is called the frequency of the signal f.
(d) The Fourier transform of the signal f(t) is also called its frequency spectrum.
(e) The magnitude of the Fourier transform ,|F (ω)|, gives the spectrum ampli-
tude, while its argument arg(F (ω)), the spectrum phase.
(f) f(t) = F−1[F (ω)] =1
2π
∫ ∞
−∞F (ω)eiωtdω is called the inverse Fourier trans-
form of F (ω).
Example 113. Find the Fourier transform of the function f(t) = e−atH(t), where
a is a positive constant and H(t) =
{0 , t < 0
1 , t ≥ 0is the unit step (or Heaviside
)function.
Answer. F (ω) =1
a + iω
Note : limt→∞
∣∣∣− e−(a+iω)t∣∣∣ = lim
t→∞e−at = 0, for a > 0 ⇒ lim
t→∞−e−(a+iω)t = 0
Example 114. Find the Fourier transform of the rectangular pulse
f(t) =
{A , |t| ≤ T
0 , |t| > T
where T is a positive constant.
Answer. F (ω) =
∫ T
−T
Ae−iωtdt = 2Asin ωT
ω
93
Reading Assignment 27. Let a be a positive constant. Find the Fourier transform
of
f(t) = e−a|t|.
You may assume the results
limt→∞
e−(a+iω)t
(a + iω)= 0 and lim
t→−∞
e(a−iω)t
(a− iω)= 0
Answer.
(a) f(t) = e−a|t| ⇒ f(t) =
{e−at , t ≥ 0
eat , t < 0
(b) The Fourier transform is
F (ω) =
∫ ∞
−∞f(t)e−iωtdt, (by definition)
=
∫ 0
−∞eate−iωtdt +
∫ ∞
0
e−ate−iωtdt
=[ e(a−iω)t
(a− iω)
]0
−∞+
[−e−(a+iω)t
(a + iω)
]∞0
=1
a− iω+
1
a + iω=
2a
a2 + ω2
That is, F{e−a|t|} =2a
a2 + ω2.
Remark 8. limt→−∞
∣∣∣ e(a−iω)t
(a− iω)
∣∣∣ = limt→−∞
∣∣∣ e(a−iω)t
(a− iω)
∣∣∣= lim
t→−∞
∣∣∣ eat
(a− iω)
∣∣∣ = 0 (since |e−iωt| = 1, a > 0 and |a− iω| 6= 0)
⇒ limt→−∞
e(a−iω)t
(a− iω)= 0
Similarly, limt→∞
e−(a+iω)t
(a + iω)= 0
94
5.3 Some Properties of the Fourier transform
Theorem 5.3.1 ( Linearity). If f and g are functions of t , then
F[af(t) + bg(t)] = aF[f(t)] + bF[g(t)]
for any constants a and b.
Theorem 5.3.2 (Time-differentiation property). F{dnf
dtn
}= (iω)nF (ω)
Proof. f(t) = F−1{F (ω)} =1
2π
∫ ∞
−∞F (ω)eiωtdω ⇒ f ′(t) =
1
2π
∫ ∞
−∞iωF (ω)eiωtdω ⇒
f ′(t) is the inverse Fourier transform of iωF (ω) ⇒ F{df
dt
}= (iω)F (ω) the result
follows by induction.
Example 115. Determine the Fourier transform of the time signal y(t) satisfying
the differential equation
y′′ + 3y′ + 4y = e−5tH(t).
Answer. Y (ω) =1
(4 + 3iω − ω2)(5 + iω)
Theorem 5.3.3 ( First Shifting Theorem (Frequency-shift property). F{eiω0tf(t)} =
F (ω − ω0)
Note This property indicates that multiplication by eiω0t shifts the spectrum of f(t)
so that it is centered on the point ω0 in the frequency domain.
Proof. F{eiω0tf(t)} =
∫ ∞
−∞eiω0tf(t)e−iωtdt =
∫ ∞
−∞f(t)e−i(ω−ω0)tdt = F (ω − ω0)
95
Example 116.
Note 21. Amplitude modulation is a technique that allows audio signals to be
transmitted as electromagnetic waves. The maximum frequency of audio signals is
typically 10 kHz. We would need to use a very large antenna to transmit these signals
directly. To overcome this problem, we will mix the audio signal (usually called the
modulation signal) with a carrier signal which has a higher frequency. As a
higher frequency signal has a shorter wavelength, we could now use a reasonable -
sized antenna.
Determine the frequency spectrum of the amplitude-modulated signal g(t) = f(t) cos ωct
if f(t) = e−2tH(t).
Hint : cos ωct =1
2(eiωct + e−iωct)
Remark 9. cos ωct is the carrier signal and f(t) is the modulation signal.
Answer.1
2
1
2 + i(ω − ωc)+
1
2
1
2 + i(ω + ωc)
Theorem 5.3.4 (Second Shifting Theorem ( Time-shift property)). F{f(t − τ)} =
e−iωτF (ω)
Remark 10. This property says that delaying a signal by a time τ causes its Fourier
transform to be multiplied by e−iωτ .
Proof. F{f(t−τ)} =
∫ ∞
−∞f(t−τ)e−iωtdt = e−iωτ
∫ ∞
−∞f(t−τ)e−iω(t−τ)dt = e−iωτ
∫ ∞
−∞f(v)e−iωvdv =
e−iωτF (ω)
Example 117. Use the result of Example 114 and the Second Shift Theorem to
determine the Fourier transform of the rectangular pulse
g(t) =
{A , 0 ≤ t ≤ 2T
0 , elsewhere
Answer. Note that the midpoint of the pulse occurs at t = T. We need to shift the
graph T units to the left to center the pulse at 0. Thus we can write g(t) = f(t− T )
where f(t) =
{A , |t| ≤ T
0 , |t| > T⇒ F{g(t)} = e−iωT F (ω) = e−iωT 2A sin ωT
ω
96
Example 118. Find the inverse Fourier transform of G(ω) =e−2iω
3 + iω.
Answer. g(t) = e−3(t−2)H(t− 2)
Theorem 5.3.5 (The symmetry (or duality) property). If f(t) and F (ω) form a
Fourier transform pair, then F (t) and 2πf(−ω) also form a Fourier transform pair
i.e., F{F (t)}(ω) = 2πf(−ω).
Proof. (a) F{F (t)}(ω) =
∫ ∞
−∞F (t)e−iωtdt, by the definition of the Fourier transform.
(b) f(t) =1
2π
∫ ∞
−∞F (ω)eiωtdω
⇒ f(ω) =1
2π
∫ ∞
−∞F (t)eitωdt, t ↔ ω
⇒ f(−ω) =1
2π
∫ ∞
−∞F (t)e−iωtdt
⇒ 2πf(−ω) =
∫ ∞
−∞F (t)e−iωtdt = F{F (t)}(ω)
97
Example 119. Given that the function f(t) =
{1 , |t| ≤ 1
0 , |t| > 1has Fourier trans-
form F (ω) = 2sin ω
ω. Use the symmetry property to deduce the Fourier transform of
the signal g(t) =sin t
t.
Answer. Given that F (ω) = 2sin ω
ω= 2g(ω).
Thus, by the symmetry property F[F (t)](ω) = 2πf(−ω).
That is, 2G(ω) = 2πf(−ω) ⇒ G(ω) = πf(−ω) =
{π , |ω| ≤ 1
0 , |ω| > 1
Example 120 (Solving ODE). Solve y′ + 3y = δ(t− 2).
[Hint:F[δ(t− a)] = e−iωa]
Answer. y = e−3(t−2)H(t− 2)
Remark 11. There is no arbitrary constant in this solution because the Fourier trans-
form has returned the only solution that is continuous and bounded for all real t.
Boundedness is assumed when we use the transform because of the required conver-
gence of
∫ ∞
−∞|y(t)|dt.
Reading Assignment 28. Use the Fourier transform to solve the differential equa-
tion
y′ − 4y = H(t)e−4t
where H is the Heaviside function.
Answer. Apply Fourier transform on the DE,
F{y′} − 4F{y} = F{H(t)e−4t}iωY − 4Y =
1
4 + iω(from table)
Solve for Y :
Y =1
(4 + iω)(iω − 4)= −1
8
2(4)
42 + ω2
y = −1
8e−4|t| (from table).
98
Example 121. Use the Fourier transform to solve the differential equation
y′′ + 3y′ + 2y = H(t)e−3t.
Answer. y =1
2H(t)e−t −H(t)e−2t +
1
2H(t)e−3t
5.4 A table of Fourier transforms
f(t) F[f(t)] = F (ω)
e−atH(t) (a > 0)1
a + iω
te−atH(t) (a > 0)1
(a + iω)2
e−a|t| (a > 0)2a
a2 + ω2
e−at2
√π
ae−ω2/4a
dnf
dtn(iω)nF (ω)
f(t− τ) e−iωτF (ω)
eiω0tf(t) F (ω − ω0)
δ(t− a) e−iωa
F (t) 2πf(−ω)
Exercise 25. Find the Fourier transform of the signal
f(t) =3
16 + t2
Answer.3π
4exp(−4|ω|)
99
100
Chapter 6
Partial Differential Equations
6.1 Revision : Half-range Expansions
Definition 6.1.1. Let f(x) be a function defined only on the finite interval (0, L).
(a) The half-range cosine series expansion of f is the cosine series
f ∗(x) = a0 +∞∑
n=1
an cosnπx
L
where a0 =1
L
∫ L
0
f(x) and an =2
L
∫ L
0
f(x) cosnπx
Ldx, n = 1, 2, . . . .
(b) The half-range sine series expansion of f is the sine series
f ∗(x) =∞∑
n=1
bn sinnπx
L
where bn =2
L
∫ L
0
f(x) sinnπx
Ldx, n = 1, 2, . . . .
101
Exercise 26. For the function f(x) = 1 defined only in the interval 0 < x < π,
obtain
(a) a half range cosine series expansion
(b) a half range sine series expansion
Answer. (a) f1∗(x) = 1. (b) f2
∗(x) =4
π
[sin x +
1
3sin 3x +
1
5sin 5x + · · ·
].
6.2 Revision : 2nd Order HLDE With Constant
Coefficients
Consider the 2nd order homogeneous linear differential equation (HLDE) with con-
stant coefficients
ay′′ + by′ + cy = 0 (1)
where a, b and c are constants.
Equation (1) has solutions eλx where λ satisfies the characteristic equation
aλ2 + bλ + c = 0.
There are 3 possible forms for the general solution of (1) depending on the roots λ1
and λ2 of the characteristic equation:
Case 1. If λ1 and λ2 are real and distinct, then the general solution of (1) is
y = c1eλ1x + c2e
λ2x.
Case 2. If λ1 = λ2 = λ are real and equal, then the general solution of (1) is
y = c1eλx + c2xeλx.
Case 3. If λ1 and λ2 are complex conjugates, say, λ1 = α+ iβ, λ2 = α− iβ, then the
general solution of (1) is
y = eαx(c1 cos βx + c2 sin βx
).
102
Exercise 27. Find a general solution to
(a) y′′ − 4y = 0
(b) y′′ − 4y′ + 4y = 0
(c) y′′ + 2y′ + 4y = 0
Answer. (a) y = c1e2x + c2e
−2x. (b) y = c1e2x + c2xe2x. (c) y = e−x[c1 cos(
√3x) +
c2 sin(√
3x)].
103
6.3 Partial Differential Equations
Definition 6.3.1. A partial differential equation is an equation that contains
partial derivatives.
Example 122. Some classical examples of PDEs.
(a) 1-D Wave Equation utt = c2uxx
(b) 1-D Heat Equation ut = c2uxx
(c) Laplace Equation uxx + uyy = 0
Definition 6.3.2. A solution of a PDE in some region R is a function that possesses
all the derivatives occurring in the equation and satisfies the equation.
Note : PDEs have a large number of solutions and general solutions are not always
readily obtained. Typically, further conditions are imposed to obtain a unique solution.
Example 123. [Some PDEs can be solved like an ODE]
(a) Find a solution u(x, y) of the PDE ux = 0.
(b) Solve for u(x, y) if uy + 2yu = 0.
104
Reading Assignment 29. Solve for u(x, y) if uxy = 0.
Answer. uxy = 0 ⇔ ∂
∂y
(∂u
∂x
)= 0.
By integrating with respect to x, we get ux = f(x).
By integrating with respect to y, we obtain the solution u(x, y) = f(x) + g(y), where
f(x) and g(y) are arbitrary functions of x and y, respectively.
Reading Assignment 30. Solve for u(x, y) if uxx − u = 0.
Answer. As the PDE contains the derivatives of one of the variables only , we can
treat it like an ODE u′′− u = 0, which has a GS of the form u = Ae−x + Bex. So the
required solution is u(x, y) = A(y)e−x+B(y)ex where A and B are arbitrary functions
of y.
Exercise 28. Verify that u(x, y) = a ln(x2+y2)+b is a solution of Laplace’s equation
and determine a and b so that u satisfies the boundary conditions u = 0 on the circle
x2 + y2 = 1 and u = 5 on the circle x2 + y2 = 9.
Answer. a = 5/(ln 9), b = 0
Definition 6.3.3. The general second-order linear partial differential equation in
two variables x and y is an equation of the form
Auxx + Buxy + Cuyy + Dux + Euy + Fu = G
where A, B, C,D,E, F and G are given functions of x and y.
A second-order linear partial differential equation in two variables x and y that is not
of the above form is a nonlinear PDE.
(a) The equation is homogeneous if G = 0 for all x and y in the domain of the
equation; otherwise, it is called nonhomogeneous
(b) If the coefficients A, B, C,D, E, and F are all constants, then the equation is
said to have constant coefficients; otherwise, it has variable coefficients.
105
Example 124. Examples of linear and nonlinear equations are
(a) ut = α2uxx (linear)
(b) uxx + uyy + u = sin x (linear)
(c) uux + yuy + u = 0 (nonlinear)
(d) uxx + uyy + u2 = 0 (nonlinear)
Definition 6.3.4. Three Basic Types of Linear Equations
The PDE Auxx + Buxy + Cuyy + Dux + Euy + Fu = G is called
(a) parabolic if B2 − 4AC = 0. Parabolic equations often describe heat flow and
diffusion phenomena, such as heat flow through the earth’s surface.
(b) hyperbolic if B2 − 4AC > 0. Hyperbolic equations often describe wave motion
and vibrating phenomena, such as violin’s strings and drumheads.
(c) elliptic if B2−4AC < 0. Elliptic equations are often used to describe steady state
phenomena and thus do not depend on time. Elliptic equations are important in
the study of electricity and magnetism.
Example 125. (a) ut − uxx = 0 is a parabolic equation since B2 − 4AC = 02 −4(−1)(0) = 0.
(b) uxx + uyy = 0 is an elliptic equation since B2 − 4AC = 02 − 4(1)(1) = −4.
(c) utt − uxx = 0 is a hyperbolic equation since B2 − 4AC = 02 − 4(−1)(1) = 4.
6.4 Solving Partial Differential Equations
One way to solve a PDE with given initial and boundary conditions is to find ”enough
” solutions of the PDE that can be ”pieced together” to produce a solution of the
PDE that also satisfies the initial and boundary conditions. A few methods that are
commonly used are
(a) Separation of Variables : This method reduces a PDE in n independent
variables to n ODEs.
106
(b) Integral Transforms : This method reduces a PDE in n independent variables
to a PDE in n− 1 independent variables. A few commonly used transforms are
Fourier, Mellin, and Hankel.
Note : In particular, a PDE in two variables would be transformed into an ODE.
After solving the ODE, we apply the inverse transform to obtain the solution to
the PDE.
(c) Numerical Methods : These methods (often) change a PDE to a system of
difference equations that can be solved by using a computer. For nonlinear equa-
tions, this is often the only technique that will provide (approximate) solutions.
Theorem 6.4.1. Superposition Principle
If u1 and u2 are any solutions of a linear homogeneous PDE in some regiond R, then
the linear combination u = c1u1 + c2u2 is also a solution of the PDE.
6.5 Eigenvalue Problems
An important class of problems known as eigenvalue problems arise in solving
PDEs using the method of separation of variables. A typical eigenvalue problem
would be to find all real values of λ, called eigenvalues , such that the boundary
value problem
X ′′ + λX = 0, 0 < x < L; X(0) = X(L) = 0
has a non-zero solution. The non-zero solution(s), if there are any, are called eigen-
functions.
Example 126. Find the eigenvalues and eigenfunctions of the problem
X ′′ + λX = 0, 0 < x < L (6.1)
X(0) = X(L) = 0 (6.2)
where λ is a parameter.
107
Answer. Consider the 3 cases:
case 1. Suppose λ < 0, say, λ = −µ2.
Then X = 0 is the only solution of the BVP, i.e. if λ < 0 we do no obtain non-trivial
solutions.
case 2. Suppose λ = 0.
case 3. Suppose λ > 0, say, λ = p2.
Then the GS of the BVP is X(x) = c1 cos px + c2 sin px.
BCs X(0) = X(L) = 0 ⇒ c1 = 0, c2 sin pL = 0.
For non-0 solutions, we need to have c2 6= 0 and sin pL = 0, i.e. pL = nπ, n =
0,±1,±2, . . .. Thus
X(x) = cn sinnπx
L, (cn arbitrary constants)
are non-0 solutions if λ is restricted tothe values λ = p2 =n2π2
L2, n = 1, 2, 3, . . .. This
is because
(i) n = 0 ⇒ p = 0, which is impossible as p > 0
(ii) negative n give the same values for λ and the same solutions since sin−nπx
L=
− sinnπx
L.
Conclusion : The eigenfunctions of the BVP are Xn(x) = sinnπx
L(put cn = 1) with
the corresponding eigenvalues λ = p2 =n2π2
L2, n = 1, 2, 3, . . ..
108
6.6 Separation Of Variables Method
Example 127. The Vibrating String Problem
Solve the initial-boundary-value problem
utt = c2uxx, 0 < x < L, t > 0 (6.3)
with boundary conditions
u(0, t) = 0, t > 0 (6.4)
u(L, t) = 0, t > 0 (6.5)
and initial conditions
u(x, 0) = f(x), 0 < x < L (6.6)
ut(x, 0) = 0, 0 < x < L (6.7)
Answer.
(i) [ Separate variables]
Substituting u(x, t) = X(x)T (t) into (6.3), we obtain two homogeneous ODEs of
X and T
X ′′ − kX = 0, X(0) = 0, X(L) = 0 (6.8)
T ′′ − c2kT = 0, T ′(0) = 0 (6.9)
where k is a constant.
(ii) [ Solve the equations (6.8) and (6.9)]
(6.9) is a second order ODE with only one IC⇒ it has non-0 solutions ∀ k.
(6.8) has non-0 solutions only when k < 0, in which case, the non-0 solutions
are
Xn = sinnπx
Lcorresonding to k = −n2π2
L2.
With these values of k, (6.9) becomes
T ′′ +n2π2
L2T = 0, T ′(0) = 0.
109
Solving this IVP, we obtain Tn = coscnπt
L.
Thus un(x, t) = Xn(x)Tn(t) = sinnπx
Lcos
cnπt
L, n = 1, 2, . . . . This satisfies all
the conditions except (6.6).
(iii) [ Solution to the entire problem ]
Let u(x, t) =∞∑
n=1
Anun(x, t) =∞∑
n=1
An sinnπx
Lcos
cnπt
L. This linear combination
also satisfies (6.3), (6.4), (6.5), and (6.7). We find An so that it also satisfies
(6.6) i.e. u(x, 0) = f(x) or u(x, 0) =∞∑
n=1
An sinnπx
L= f(x) i.e. An are the
Fourier coefficients of the half range expansion of f .
Reading Assignment 31. Use the method of separation of variables to solve the
boundary value problem
ut = uxx, 0 < x < π, t > 0 (6.10)
u(0, t) = 0, t > 0 (6.11)
ux(π, t) = 0, t > 0 (6.12)
u(x, 0) = 3 sin1
2x− sin
5
2x, 0 ≤ x ≤ π (6.13)
Record all your arguments in detail. You may assume the result that the eigenvalues
of the eigenvalue problem
X ′′ − kX = 0, X(0) = X ′(π) = 0
are
k = −(2n− 1)2/4, n = 1, 2, . . .
and that
Xn = sin(2n− 1)x
2, n = 1, 2, . . .
are the corresponding eigenfunctions.
Answer. Sub u(x, t) = X(x)T (t) into (6.10):
XT ′ = X ′′T ⇒ X ′′
X=
T ′
T= k
110
⇒ X ′′ − kX = 0, T ′ − kT = 0
(6.11) and (6.12) ⇒ X(0) = X ′(π) = 0 ⇒
X ′′ − kX = 0, X(0) = X ′(π) = 0 (6.14)
T ′ − kT = 0 (6.15)
With k = −(2n− 1)2/4, (6.15)⇒ T ′ +(2n− 1)2
4T = 0
⇒ Tn = exp[− (2n− 1)2
4t]
Let u(x, t) =∞∑
n=1
An sin(2n− 1)x
2exp
[− (2n− 1)2
4t].
(6.13)⇒∞∑
n=1
An sin(2n− 1)x
2= 3 sin
1
2x − sin
5
2x ⇒ A1 = 3, A3 = −1, and other
An = 0.
The solution is u(x, t) = 3e−t/4 sin1
2x− e−25t/4 sin
5
2x.
Exercise 29. Solve the boundary-value problem
ut = uxx, 0 < x < π, t > 0
ux(0, t) = ux(π, t) = 0, t > 0
u(x, 0) = 3− sin2 x, 0 ≤ x ≤ π
Note : This BVP can be thought of as a model of heat conduction in a bar of length
π. The bar is insulated at both ends. In this case, the solution u(x, t) represents the
temperature of the bar when the initial temperature is 3− sin2 x.
Answer.5
2+
1
2e−4t cos(2x).
111
Reading Assignment 32. Solve the PDE ut = 4uxx by the method of separation of
variables.
Answer. Using u(x, t) = X(x)T (t),
we obtain XT ′ = 4X ′′T orX ′′
X=
T ′
4T= k, a constant.
That is, we obtain the equations X ′′ − kX = 0, T ′ − 4kT = 0.
We consider the following 3 cases:
(a) If k = 0, then we obtain X ′′ = 0 and T ′ = 0.
Integrating, we obtain X = c1 + c2x and T = c3.
Hence, the solution u(x, t) = c3(c1 + c2x) = A + Bx, where A = c1c3 and B = c2c3
are arbitrary constants.
(b) If k = p2 > 0, we obtain X ′′ − p2X = 0 and T ′ − 4p2T = 0.
The solutions for these two equations are
X = c1epx + c2e
−px and T = c3e4p2t.
Hence, u(x, t) = [c1epx + c2e
−px]c3e4p2t = [Aepx + Be−px]e4p2t.
(c) If k = −p2 < 0, we obtain X ′′ − p2X = 0 and T ′ + 4p2T = 0.
The solutions for these two equations are
X = c1 cos px + c2 sin px and T = c3e−4p2t.
Hence, u(x, t) = [c1 cos px + c2 sin px]c3e−4p2t = [A cos px + B sin px]e4p2t.
Exercise 30. Solve the PDE by the method of separation of variables.
(a) ut = 4uxx (b) yux + xuy = 0
Answer. (a) k = 0 : u = Ax + B; k = p2 > 0 : u = [Aepx + Be−px]e4p2t; k = −p2 <
0 : u = [A cos px + B sin px]e4p2t (b) Aek(x2−y2)/2
112
Bibliography
[1] B.V. Ramana, Engineering Mathematics.
[2] Advanced Modern Engineering Mathematics, Glyn James
[3] Erwin Kreyszig, Advanced Engineering Mathematics.
[4] K.A. Stroud, Advanced Engineering Mathematics.
[5] Anthony Croft, Robert Davison, Martin Hargreaves,Engineering Mathematics.
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