Lecture Notes for
MATH2230
Neil Ramsamooj
Table of contents
1 Vector Calculus
................................................ 5
1.1 Parametric curves and arc length . . . . . . . . . . . . .
1.2 Review of partial differentiation . . . . . . . . . . . . . .
1.3 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Divergence and curl of a vector field . . . . . . . .
1.3.2 Gradient of a function . . . . . . . . . . . . . . . . .
1.4 Line integrals and double integrals . . . . . . . . . . . .
1.4.1 Line integrals . . . . . . . . . . . . . . . . . . . . . .
1.4.2 Path independence and conservative vector fields
1.4.3 Double integrals . . . . . . . . . . . . . . . . . . . .
1.5 Green’s theorem . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Surface integrals . . . . . . . . . . . . . . . . . . . . . . . .
1.6.1 Parametric surfaces . . . . . . . . . . . . . . . . . .
1.6.2 Surface integrals . . . . . . . . . . . . . . . . . . . .
1.6.3 Surface integrals over vector fields . . . . . . . . .
1.7 Triple integrals and Divergence theorem . . . . . . . . .
1.7.1 Triple integrals . . . . . . . . . . . . . . . . . . . . .
1.7.2 Divergence theorem . . . . . . . . . . . . . . . . . .
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.. 5
.. 9
. 14
. 14
. 17
. 23
. 23
. 25
. 30
. 37
. 39
. 39
. 42
. 48
. 56
. 56
. 58
2 Laplace transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
2.1 Definition and existence of Laplace transforms . . . . . . . . .
2.1.1 Improper integrals . . . . . . . . . . . . . . . . . . . . . . .
2.1.2 Definition and examples of Laplace tranform . . . . . .
2.1.3 Existence of Laplace transform . . . . . . . . . . . . . . .
2.2 Properties of Laplace transforms . . . . . . . . . . . . . . . . .
2.2.1 Linearity property . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 The inverse Laplace transform . . . . . . . . . . . . . . .
2.2.3 Shifting the s variable; shifting the t variable . . . . . .
2.2.4 Laplace transform of derivatives . . . . . . . . . . . . . .
2.3 Applications and more properties of Laplace transforms . .
2.3.1 Solving differential equations using Laplace transforms
2.3.2 Solving simultaneous linear differential equations using
the Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.3 Convolution and Integral equations . . . . . . . . . . . .
2.3.4 Dirac’s delta function . . . . . . . . . . . . . . . . . . . . .
2.3.5 Differentiation of transforms . . . . . . . . . . . . . . . . .
2.3.6 The Gamma function Γ(x) . . . . . . . . . . . . . . . . . .
67
67
69
73
78
78
78
82
87
89
89
3 Fourier series
3.1
3.2
3.3
3.4
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. . 93
. . 95
. . 97
. 105
. 106
................................................
Definitions . . . . . . . . . . . . . . .
Convergence of Fourier Series . . .
Even and odd functions . . . . . . .
Half range expansions . . . . . . . .
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111
114
121
126
....................................
131
4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 The Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1 A derivation of the heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
131
135
135
3
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111
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4 Partial Differential Equations
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4
4.2.2 Solution of the heat equation by seperation of variables . . . . .
4.2.3 The heat equation with insulated ends as boundary conditions
4.2.4 The heat equation with nonhomogeneous boundary conditions
4.3 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.1 A derivation of the wave equation . . . . . . . . . . . . . . . . . .
4.3.2 Solution of the wave equation by seperation of variables . . . .
4.4 Laplace’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.1 Solving Laplace’s equation by seperation of variables . . . . . .
4.5 Laplace’s equation in polar coordinates . . . . . . . . . . . . . . . . . .
Table of contents
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138
143
148
154
154
157
163
166
171
Chapter 1
Vector Calculus
1.1 Parametric curves and arc length
Recall that a function y = f (x) describes a curve in the Cartesian plane which consists of points
(x, f (x))
where x is the independent variable. Another method of defining a curve in the Cartesian plane is by
the use of a parameter t.
Definition 1.1. A parametric curve C in the Cartesian plane is obtained by specifying x and y
to be functions of a parameter t
x = f (t)
y = g(t)
where a 6 t 6 b.
Notice that any specific value of the parameter t = t0 describes a particular point (x(t0), y(t0)) of C;
the curve C consists of the set of such points
C = {(x(t), y(t))|a 6 t 6 b}.
Example 1.2. Sketch the parametric curve C
π
x = cos t
y = sin t
where 0 6 t 6 4 .
Answer: From the trigonometric identity
cos2t + sin2t = 1
we have
x2 + y 2 = 1,
and therefore the curve C consists of points (x, y) that lie on a circle of radius 1. The values of the
π
π
parameter t = 0, t = 4 at the ends of the interval 0 6 t 6 4 correspond respectively to the points (1,
1
1
0) and ( √ , √ ).
2
2
1
Therefore the curve C consists of the points on the circle of radius 1 that lie between (1, 0) and ( √ ,
2
1
√ ):
2
5
6
Vector Calculus
Notation 1.3. A parametric curve
x = f (t)
y = g(t)
where a 6 t 6 b can also be written in vector notation
r(t) = f (t)i + g(t)j
where
a6t6b
where
a6t6b
where i and j are the standard unit vectors.
Definition 1.4. A parametric curve
r(t) = f (t)i + g(t)j
is said to be smooth if the component functions f (t) and g(t) have continuous derivatives f ′(t),
g ′(t) respectively on the interval a < t < b.
The definition of a parametric curve in three dimensional space is analogous to the definition in the
Cartestian plane.
Definition 1.5.
A parametric curve C in the three dimensional space is obtained by specifying x, y and z to
be continuous functions of a parameter t
x = f (t)
y = g(t)
z = h(t)
where a 6 t 6 b. Such a parametric curve C can also be written in vector notation
r(t) = f (t)i + g(t)j + h(t)k where
a6t6b
where i, j and k are the standard unit vectors. Furthermore, C is said to be smooth if the component functions f (t), g(t) and h(t) have continuous derivatives f ′(t), g ′(t) and h ′(t) respectively on the
interval a < t < b.
Example 1.6. Sketch the parametric curve
r(t) = ti + tj + (1 − t)k
where
0 6 t 6 1.
Answer: Notice that each of the functions x, y and z are linear functions of t
x=t
y=t
z=1−t
so the curve C will be a segment of a straight line. The values of the parameter t = 0, t = 1 at the
ends of the interval 0 6 t 6 1 correspond respectively to the points (0, 0, 1) and (1, 1, 0). Therefore
the curve C consists of the line that connects the points (0, 0, 1) and (1, 1, 0):
7
1.1 Parametric curves and arc length
The following results express the arc length of a parametric curve as an integral.
Theorem 1.7.
i. The arc length s of a smooth parametric curve C
r(t) = x(t)i + y(t)j
where
a6t6b
in the Cartesian plane is given by
s=
Z
s
b
a
dx
dt
2
+
dy
dt
2
dt.
ii. The arc length s of a smooth parametric curve C
r(t) = x(t)i + y(t)j + z(t)k
in three dimensional space is given by
Z s
b
s=
a
dx
dt
2
+
dy
dt
where
2
a6t6b
dz
dt
2
06t6
π
4
+
Example 1.8. Find the arc length of the parametric curve
r(t) = cos ti + sin tj
where
of Example 1.2.
Answer:
s=
Z
b
a
=
Z
0
=
Z
0
=
Z
0
π
4
π
4
π
4
π
4
= [t]0
π
= .
4
s
dx
dt
2
+
dy
dt
2
dt
q
2
2
( − sin t) + (cos t) dt
√
sin2 t + cos2 t dt
√
1 dt
dt.
8
Vector Calculus
Notice from the diagram in Example 1.2, this answer agrees with the formula for the arc length of a
circle
s = rθ
π
= (1)
4
π
= .
4
9
1.2 Review of partial differentiation
1.2 Review of partial differentiation
Recall for a function of one variable f (x), the derivative at x = a is given by
f (a + h) − f (a)
h
h→0
f ′(a) = lim
(1.1)
and f ′(a) has the geometric interpretation as the slope of the tangent line to f (x) at x = a as may
be seen in the following diagram.
Example 1.9. If f (x) = x2 then f ′(x) = 2x. At the value x = 1, we have f ′(1) = 2 and so the slope
of the tangent to the point (1, 1) is equal to 2 as illustrated in the following diagram
The limit definition (1.1) is not usually used to compute derivatives, in practice derivatives are
computed by a set of rules – power rule, product rule, quotient rule etc. Similarly, partial derivatives are defined using limits, but actually computed by using rules. The following are the definitions of partial derivatives of a function f (x, y) of two variables
Definition 1.10.
i. If f (x, y) is a function of two variables then the partial derivative of f with respect to
∂f
x at the point (a, b) is denoted as fx(a, b) or ∂x (a, b) and is defined as
fx(a, b) = lim
h→0
f (a + h, b) − f (a, b)
h
ii. If f (x, y) is a function of two variables then the partial derivative of f with respect to
∂f
y at the point (a, b) is denoted as f y (a, b) or ∂y (a, b) and is defined as
f (a, b + h) − f (a, b)
h
h→0
f y(a, b) = lim
Partial derivatives of a function of two variables are usually computed by the following
Rule for finding partial derivatives of f (x, y)
i. To find fx, treat y as a constant and differentiate with respect to x.
ii. To find f y , treat x as a constant and differentiate with respect to y.
∂f
∂f
Example 1.11. Let f (x, y) = x2 y 3. Find ∂x and ∂y .
∂f
Answer: To find ∂x , treat y as a constant in f (x, y) = x2 y3 . One can imagine that y is equal to
′′
some fixed constant, say y = 11. Therefore ′′ f (x, y) = x2113 and now differentiate x2113 as usual
with respect to x to get (x2113) ′ = 2x113. Now replace the 11 by y to get the answer
∂f
= 2xy3.
∂x
10
Vector Calculus
∂f
Similarly to find ∂y , treat x as a constant in f (x, y) = x2 y3. As before, imagine that x is equal to a
fixed constant, say x = 7. Therefore f (x, y) = 72 y 3 ′′ and now differentiate 72 y 3 as usual with respect
to y to get (72 y 3) ′ = 72 (3y 2). Now replace the 7 by x to get the answer
∂f
= x2(3 y 2) = 3x2 y 2.
∂y
∂f
∂f
Example 1.12. Let f (x, y) = x2cos(x + y). Find ∂x and ∂y .
∂f
Answer: To find ∂x , treat y as a constant in f (x, y) = x2cos(x + y), say y = 13. Then ′′ f (x, y) =
x2cos(x + 13) ′′ and this is a product of two functions, therefore we use the usual product rule to differentiate with respect to x
(x2cos(x + 13)) ′ = (x2) ′cos(x + 13) + x2(cos(x + 13) ′)
= 2x cos(x + 13) − x2sin(x + 13)
and replacing the 13 by y we have the answer
∂f
= 2x cos(x + y) − x2sin(x + y).
∂x
∂f
To find ∂y , treat x as a constant. In this case f (x, y) = x2cos(x + y) is a product of a constant x2
and a function cos(x + y) and so it is not necessary to use product rule. We have the answer
∂f
∂
= x2 (cos(x + y))
∂y
∂y
= x2( − sin(x + y))
= − x2sin(x + y).
d2 f
Recall that one obtains the second derivative dx2 of a function f (x) of one variable by differentiating the first derivative, for example
f (x) = sin x
df
⇒
= cos x
dx
2
d df
d f
⇒ 2=
dx dx
dx
d
=
(cos x)
dx
= − sin x.
∂2f
∂2f
∂2f
∂2f
Similarly, one obtains the second partial derivatives ∂x2 , ∂y 2 , ∂x∂y and ∂y∂x of a function f (x,
∂f
∂f
y) of two variables by taking the partial derivative of a first partial derivative ∂x or ∂y . To be precise
∂2f
∂ ∂f
=
∂x2
∂x ∂x
∂2f
∂ ∂f
=
∂y 2 ∂y ∂y
∂2f
∂ ∂f
=
∂y∂x ∂y ∂x
∂ ∂f
∂2f
=
.
∂x∂y ∂x ∂y
11
1.2 Review of partial differentiation
Example 1.13. Find the second partial derivatives of f (x, y) = x2 y + y 3.
Answer: The first partial derivatives are
∂f
= 2xy
∂x
∂f
= x2 + 3y 2
∂y
∂2f
∂f
∂2f
∂f
∂2f
∂f
∂2f
∂f
The second partial derivative ∂x2 is obtained by taking the partial derivative of ∂x with respect to
x
∂2f
∂ ∂f
=
∂x2
∂x ∂x
∂
= (2xy)
∂x
= 2y
The second partial derivative ∂ 2 y is obtained by taking the partial derivative of ∂y with respect to
y
∂ ∂f
∂2f
=
∂y ∂y
∂y 2
∂ 2
x + 3y2
=
∂y
= 6y
The second partial derivative ∂x∂y is obtained by taking the partial derivative of ∂y with respect to
x
∂2f
∂ ∂f
=
∂x∂y ∂x ∂y
∂ 2
=
x + 3y 2
∂x
= 2x
The second partial derivative ∂y∂x is obtained by taking the partial derivative of ∂x with respect to
y
∂ ∂f
∂2f
=
∂y∂x ∂y ∂x
∂
= (2xy)
∂y
= 2x
The partial derivatives of a function f (x, y, z) of three variables are defined in a similar manner to
∂f
the partial derivatives of a function of two variables – for example, to obtain ∂y , one treats the variables x and z as constants and differentiates with respect to y.
Example 1.14. Find the first and second partial derivatives of f (x, y, z) = x2 yz + xez.
Answer:
∂f
= 2xyz + ez
∂x
∂2f
= 2yz
∂x2
∂2f
= 2xz
∂y∂x
∂2f
= 2xy + ez
∂z∂x
∂f
= x2z
∂y
∂2f
=0
∂y 2
∂2f
= 2xz
∂x∂y
∂2f
= x2
∂z∂y
∂f
= x2 y + xez
∂z
∂2f
= xez
∂z 2
∂2f
= 2xy + ez
∂x∂z
∂2f
= x2
∂y∂z
12
Vector Calculus
We saw in Example 1.9 that the derivative f ′(a) can be interpreted as the slope of the tangent of
the function f (x) at the point x = a. In a similar manner, if f (x, y) is a function of two variables
∂f
∂f
then the first partial derivatives ∂x (a, b) and ∂y (a, b) may also be interpreted as slopes of lines
passing through the point corresponding to (x, y) = (a, b).
Recall that the graph of a function f (x, y) is obtained by plotting points (x, y, f (x, y)) in three–
dimensional space. In this way, the graph of a function of two variables forms a surface in 3 dimensions. For example, the graph of the function f (x, y) = 1 − x forms a surface that is a plane:
∂f
∂f
We illustrate the geometric interpretation of the partial derivatives ∂x , ∂y by considering the
1
behaviour of the function f (x, y) = 1 − x at (x, y) = ( 2 , 0). By taking partial derivatives, we have
that
∂f 1
∂f 1
,0 =−1
, 0 = 0.
∂y 2
∂x 2
1
1
1
Now notice that f ( 2 , 0) = 1 − 2 = 2 . Therefore the point
1
1
1
1
, 0, f ( , 0) =
, 0,
2
2
2
2
1
1
lies on the graph of f (x, y) = 1 − x as we see in the above diagram. The
z
−
coordinate
of
,
0,
2
2
1
1
1
is the function value f ( 2 , 0) and corresponds to the height of the point 2 , 0, f ( 2 , 0) above the xyplane.
1
1
We now keep x = 2 fixed and change the y-value. Plotting such points f ( 2 , y) gives the line B on
the above diagram. Notice from the diagram that line B is parallel to the x y-plane, that is, the
1
value of the function f does not change if we keep x = 2 fixed and change the y-value. This corresponds to
∂f 1
,0 =0
∂y 2
1.2 Review of partial differentiation
13
∂f 1
because the partial derivative ∂y ( 2 , 0) measures the rate at which at the function f is increasing as
1
the y-value changes while keeping x = 2 fixed.
Now keep y = 0 fixed and change the x-value. Plotting such points f (x, 0) gives the line A in the
above diagram. Notice that the height of the points on line A is decreasing as the x-values increase.
This corresponds to
∂f 1
,0 =−1
∂x 2
because the partial derivative measures the rate at which the function f is increasing as the x-value
increases while keeping y = 0 fixed.
14
Vector Calculus
1.3 Vector Fields
Definition 1.15.
A vector field on the Cartesian plane R2 is a function F that assigns a two dimensional vector
F (x, y) to each point (x, y). We may write F in terms of its component functions
F (x, y) = P (x, y)i + Q(x, y)j.
Furthermore, the vector field F is said to be continuous if and only if each its component functions
is continuous.
Example 1.16. The following diagram is a plot of the vector field
F (x, y) = − yi + xj
on the Cartesian plane. Notice that each point (x, y) on the Cartesian plane has a vector associated
to it.
Similarly, we may define vector fields in the three dimensional space R3.
Definition 1.17. A vector field on the three dimensional space R3 is a function F that assigns a
three dimensional vector F (x, y, z) to each point (x, y, z). We may write F in terms of its component functions
F (x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k.
Furthermore, the vector field F is said to be continuous if and only if each its component functions
is continuous.
1.3.1 Divergence and curl of a vector field
Definition 1.18. Let
F (x, y) = P (x, y)i + Q(x, y)j
be a vector field on the Cartesian plane R2 where the first partial derivatives of the component functions P and Q exist. Then the divergence of the vector field F is the function
div(F ) =
∂P ∂Q
+
.
∂x
∂y
15
1.3 Vector Fields
Example 1.19. Find the divergence of the following vector fields on the Cartesian plane R2
i. F (x, y) = 3i
ii. G(x, y) = 3x2i
Answer:
i.
∂P ∂Q
+
∂x ∂x
∂
∂
= (3) + (0)
∂x
∂x
=0
div(F ) =
ii.
∂P ∂Q
+
∂x ∂x
∂
∂
= (3x2) + (0)
∂x
∂x
= 6x
div(G) =
The divergence of a vector field has the following interpretation. Consider a infinitesimally small box
of length ∆x and width ∆y located at the point (x, y)
Then the divergence of a vector field F at the point (x, y) can be viewed as the net flow of F out of
an infinitesimally small box located at the point (x, y).
Consider the two vector fields F and G of Example 1.19. Notice that the vector field F (x, y) = 3i
is a constant vector field
so the net flow of F out of the small box located at the point (x, y) is zero, which agrees with the
answer div(F ) = 0 obtained in Example 1.19.
16
Vector Calculus
Notice however that the vector field G(x, y) = 3x2i is nonconstant, and that the length of the vectors
increases as x increases. From the diagram below
we see that the vectors at one vertical side of the small box are longer than the vectors at the other
vertical side. Hence one can regard the net flow out of the box as positive, which agrees with the
answer div(G) = 6x1.1 obtained in Example 1.19.
The following is the definition of the divergence of a three dimensional vector field.
Definition 1.20. Let
F (x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k
be a vector field on three dimensional space R3 where the first partial derivatives of the component
functions P , Q and R exist. Then the divergence of the vector field F is the function
div(F ) =
∂P ∂Q ∂R
+
+
.
∂x
∂y
∂z
The divergence of a three dimensional vector field can be expressed in an alternative form by the use
of a differential operator
Definition 1.21. Let ∇ denote the vector differential operator
∇=
∂
∂
∂
i + j + k.
∂x
∂y
∂z
Then the divergence of the vector field
F (x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k
is given by the dot product of ∇ and F
div(F ) = ∇.F
It is not hard to check that Definition 1.20 and Definition 1.21 are equivalent:
div(F ) = ∇.F
∂
∂
∂
=
i + j + k · (P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k)
∂x
∂y
∂z
∂
∂
∂
= (P ) + (Q) + (R)
∂y
∂z
∂x
∂P ∂Q ∂R
=
+
+
.
∂x
∂y
∂z
1.1. For a box located at (x, y) where x > 0. If the box is located on the opposite side of the y − axis, then the vectors
reverse direction and our interpretation still holds.
17
1.3 Vector Fields
Note that the divergence of three dimensional vector field has a similar interpretation to the divergence in two dimensions – div(F ) at the point (x, y, z) can be viewed as the net flow of F out of an
infinitesimally small cube located at the point (x, y, z).
Definition 1.22. Let
F (x, y, z) = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k
be a vector field on three dimensional space R3 where the first partial derivatives of the component
functions P , Q and R exist. Then the curl of the vector field F is the vector field on R3 defined by
curl (F ) = ∇ × F
i j k ∂ ∂ ∂ = ∂x ∂y ∂z P Q R ∂P ∂R
∂Q ∂P
∂R ∂Q
−
−
−
i+
j+
k.
=
∂z
∂x
∂y
∂z
∂x
∂y
Example 1.23. Find the curl of the vector field F (x, y, z) = xzi + xyzj − y2k.
Answer:
curl (F ) = ∇ × F
j
k
i
∂
∂
∂
= ∂x ∂y
∂z
xz xyz − y 2
= ( − 2y − xy)i + (x − 0)j + (yz − 0)k
= ( − 2y − xy)i + xj + yzk.
Note that at any specific point (x, y, z), curl (F ) is a three dimensional vector and so corresponds to
some specific direction in three dimensional space R3. Consider a small paddle that can rotate along
an axis. If the vector field F is viewed as the velocity of a liquid then curl (F ) at the point (x, y, z)
would be the direction in which the axis of the paddle should be aligned in order to get the fastest
counterclockwise rotation at the point (x, y, z) .
1.3.2 Gradient of a function
Definition 1.24. If f (x, y) is a function of two variables defined on the Cartesian plane R2 where
the first partial derivatives of f (x, y) exist, then the gradient of the function, denoted as grad(f ), is
the vector function defined as
∂f
∂f
grad(f ) = i + j.
∂y
∂x
Remark 1.25.
Note that grad(f ) assigns a two dimensional vector
∂f
∂f
(x0, y0)i + (x0, y0)j
∂y
∂x
to each point (x0, y0) of the Cartesian plane R2. It follows from Definition 1.15 that grad(f ) is a
vector field and because of this, grad(f ) is also referred to as the gradient vector field of f.
18
Vector Calculus
Example 1.26. Let f (x, y) = x2 + y 2. Then
∂
∂ 2
(x + y 2)i + (x2 + y 2)j
∂y
∂x
= 2xi + 2y j.
grad(f ) =
At any specific point (x, y), grad(f ) is a vector that gives the direction of the maximum increase of
the function f (x, y). Consider the following contour plot of the function f (x, y) = x2 + y 2
Figure 1.1. Contour plot of f (x, y) = x2 + y 2
Note that all points (x, y) on the circle labelled f = 1 have function value equal to 1. The point
√
(x, y) = (1/2, 3 /2) is one such point on the circle f = 1. Notice that if we change the (x, y) values
in the direction of arrow A, that is if we take (x, y) values along the circle f = 1, then the function
f (x, y) does not change value. In order to get the maximum increase in the function f (x, y) we must
move in the direction perpendicular to the circle f = 1; this is the direction specified by
grad(f ) = 2xi + 2y j
√
= i + 3j
in the diagram.
Definition 1.27. Let (x, y) be a point on a curve C in the Cartesian plane R2 that has a welldefined tangent vector T . Then any vector n at the point (x, y) that is perpendicular to the tangent
vector T is called a normal vector to the curve C at the point (x, y).
Example 1.28.
√ In Figure 1.1 above, the arrow A is a tangent vector to the curve f = 1 at the point
(x, y) = (1/2, 3 /2). The vector
√
grad(f ) = i + 3 j
√
is an example of a normal vector to the curve f = 1 at the point (1/2, 3 /2).
Definition 1.29. Let f (x, y) be a function of two variables and let c be a constant. The set of
points that satisfy
f (x, y) = c
called a level curve of the function f (x, y).
19
1.3 Vector Fields
Example 1.30. Let f (x, y) = x2 + y 2. Then the level curve
f (x, y) = 1
is the circle
x2 + y2 = 1.
Notice that the three level curves f (x, y) = 1, f (x, y) = .49 and f (x, y) = .25 are shown in Figure 1.1
above.
Theorem 1.31. Let f (x, y) = c be a level curve C in the Cartesian plane where f (x, y) is a differentiable function. Then if grad(f ) at a point (x, y) is not the zero vector, then it is a normal vector
to the curve C at the point (x, y).
Example 1.32. Find a normal vector to the ellipse x2 + 4y 2 = 1 at the point
1
1
√ , √
2 2 2
.
Answer: Notice that the curve x2 + 4y 2 = 1 is a level curve as it is in the form
f (x, y) = 1
where f (x, y) = x2 + 4y2. The gradient of the function f is
∂f
∂f
i+ j
∂y
∂x
= 2xi + 8yj
grad(f ) =
and at the point (x, y) =
1
1
√ , √
2 2 2
, grad(f ) is equal to
grad(f ) = 2xi + 8yj
2
8
= √ i+ √ j
2√ 2
√2
= 2i + 2 2 j
which, by Theorem 1.31 is a normal vector to x2 + 4y 2 = 1 at the point
1
1
√ , √
2 2 2
.
∂f
Recall that the partial derivative ∂x (x0, y0) measures the rate at which the function f is increasing
∂f
as the x-value increases while keeping the y − value fixed, in other words, ∂x (x0, y0) gives the rate of
change in the function f as the (x, y) points move from (x0, y0) in the direction of the horizontal
unit vector i; see Figure 1.2 below.
∂f
Similarly the partial derivative ∂y (x0, y0) gives the rate of change in the function f as the (x, y)
points move from (x0, y0) in the direction of the vertical unit vector j as is also illustrated in Figure
1.2 below.
20
Vector Calculus
Figure 1.2. Diagram of the domain of the function f
If we wish to determine the rate at which a function changes as the (x, y) points move from a fixed
point (x0, y0) in the direction of a unit vector that is not horizontal or vertical, then we shall need
the following definition.
Definition 1.33. The directional derivative of a function f (x, y) in the direction of a unit
vector u at a point (x0, y0), denoted as Duf , is defined as the dot product
Duf = u · grad(f )
Example 1.34. Find the directional derivative of the function f (x, y) = x2 + y 2 at the point (1, 2)
in the direction i − j.
Answer: In the definition of the directional derivative, the direction is specified by a unit vector. We
first find the unit vector u parallel to i − j
i−j
length of i − j
i−j
=p
2
1 + ( − 1)2
1
1
= √ i− √ j
2
2
u=
The vector grad(f ) at the point (x, y) = (1, 2) is
∂f
∂f
i+ j
∂x
∂y
= 2xi + 2yj
= 2i + 4j
grad(f ) =
and the required directional derivative is
Duf = u · grad(f )
1
1
= √ i − √ j · (2i + 4j)
2
√2
= − 2.
21
1.3 Vector Fields
The gradient of a function f (x, y, z) of three variables is defined and has properties analogous to the
gradient of a function of two variables.
Definition 1.35. If f (x, y, z) is a function of three variables, then the gradient of the function,
denoted as grad(f ), is the vector function defined as
∂f
∂f
∂f
grad(f ) = i + j + k.
∂x
∂y
∂z
Note that the vector differential operator of Definition 1.21 can be used to give an alternative
expression of grad(f )
grad(f ) = ∇f .
Definition 1.36. Let f (x, y, z) be a function of three variabless and let c be a constant. The set of
points that satisfy
f (x, y, z) = c
called a level surface of the function f (x, y, z).
Example 1.37. Let f (x, y, z) = x2 + y 2 + z 2. Then the level surface S
f (x, y, z) = 1
is the sphere
x2 + y 2 + z 2 = 1.
of radius 1 is illustrated in Figure 1.3 below.
Figure 1.3.
Definition 1.38. The tangent plane to a point P on a surface S is the plane that touches the
surface at the point P . A normal vector to a surface S at the point P is a vector that is perpendicular to the tangent plane at the point P .
Theorem 1.39. Let f (x, y, z) = a be a surface S in three dimensional space where f (x, y, z) is a
differentiable function. Then if grad(f ) at a point (x, y, z) is not the zero vector, then it is a normal
vector to the surface S at the point (x, y, z).
22
Vector Calculus
Example 1.40. Find a normal vector to the sphere x2 + y 2 + z 2 = 1 at the point (1, 0, 0).
Answer: Notice that the surface x2 + y 2 + z 2 = 1 is in the form
f (x, y, z) = 1
where f (x, y, z) = x2 + y 2 + z 2. The gradient of the function f is
∂f
∂f
∂f
i+ j + k
∂x
∂y
∂z
= 2xi + 2yj + 2zk
grad(f ) =
and at the point (x, y) = (1, 0, 0), grad(f ) is equal to
grad(f ) = 2xi + 2yj + 2zk
= 2i + 0j + 0k
= 2i
which, by Theorem 1.39 is a normal vector to the sphere x2 + y 2 + z 2 = 1 at the point (1, 0, 0).
∂f
∂f
As in the case of a function of two variables, the partial derivatives ∂x (x0, y0, z0), ∂y (x0, y0, z0) and
∂f
(x , y , z ) measure the rate at which a function f (x, y, z) changes as the (x, y, z) points move
∂z 0 0 0
from a fixed point (x0, y0, z0) in the direction of a unit vectors i, j and k respectively.
If we wish to determine the rate at which a function changes as the (x, y, z) points move from a
fixed point (x0, y0, z0) in the direction of a unit vector that is not a standard unit vector i, j or k,
then, as in the case of a function of two variables, we need the definition of a directional derivative.
Definition 1.41. The directional derivative of a function f (x, y, z) in the direction of a unit
vector u at a point (x0, y0, z0), denoted as Duf , is defined as the dot product
Duf = u · grad(f ).
23
1.4 Line integrals and double integrals
1.4 Line integrals and double integrals
1.4.1 Line integrals
Definition 1.42. Let F (x, y) be a continuous vector field defined on the Cartesian plane and let C
be the smooth parametric curve
r(t) = x(t)i + y(t)j
a 6 t 6 b.
Then the line integral of the vector field F along the curve C is
Z b
Z
F · dr =
F (x(t), y(t)) · (x ′(t)i + y ′(t)j ) dt
a
C
R
A line integral C F · dr can be interpreted as the work done on a particle by a force field F as it
travels on a path C. Consider the following example.
Example 1.43. Let F (x, y) = 3x2i be a vector field on the Cartesian plane. Calculate the following
line integrals
R
i. C1 F · dr where C1 is the curve
r(t) = (1 + t)i + j
ii.
R
C2
06t61
F · dr where C2 is the curve
r(t) = i + (1 + t)j
0 6 t 6 1.
Answer:
i.
Z
C1
F · dr =
=
Z
a
Z
0
=
b
Z
1
1
Z0 1
F (x(t), y(t)) · (x ′(t)i + y ′(t)j ) dt
F ((1 + t), 1) · ((1 + t) ′i + (1) ′ j ) dt
3(1 + t)2i · (1i + 0j ) dt
3(1 + t)2 dt
0
1
= (1 + t)3 0
= 8.
=
ii.
Z
C2
F · dr =
=
=
=
=
Z
b
a
Z
Z0
1
1
Z0 1
Z0 1
0
= 0.
F (x(t), y(t)) · (x ′(t)i + y ′(t)j ) dt
F (1, (1 + t)) · ((1) ′i + (1 + t) ′ j ) dt
3(1)2i · (0i + 1j ) dt
3i · (0i + 1j ) dt
0dt
24
Vector Calculus
Figure 1.4 below is an illustration of the vector field F (x, y) = 3x2i and the two curves
C1: r(t) = (1 + t)i + j
06t61
C2: r(t) = i + (1 + t)j
0 6 t 6 1.
If we regard the vector field F as a force field then as C1 lies in the direction of F , we expect F to
add energy to a particle travelling along the path C1. This agrees with
Z
F · dr = 8
C1
in part i). However, the curve C2 is perpendicular to the direction of vector field and therefore F
does not add any energy to a particle travelling along C2 which agrees with the answer
Z
F · dr = 0.
C2
in part ii).
Figure 1.4.
The definition of a line integral in three dimensional space is similar to the definition in the Cartesian plane.
Definition 1.44. Let F (x, y, z) be a continuous vector field defined in three dimensional space and
let C be the smooth parametric curve
r(t) = x(t)i + y(t)j + z(t)k
a 6 t 6 b.
Then the line integral of the vector field F along the curve C is
Z
Z b
F · dr =
F (x(t), y(t), z(t)) · (x ′(t)i + y ′(t)j + z ′(t)k ) dt.
C
a
Example 1.45. Evaluate
Z
C
F · dr
where F is the vector field in three dimensional space
F (x, y, z) = xyi + yzj + zxk
and C is the parametric curve
r(t) = ti + t2 j + t3k
0 6 t 6 1.
25
1.4 Line integrals and double integrals
Answer: From Definition 1.44 above, we have
Z b
Z
F · dr =
F (x(t), y(t), z(t)) · (x ′(t)i + y ′(t)j + z ′(t)) dt
a
C
Z 1
=
F (t, t2, t3) · (t) ′i + (t2) ′ j + (t3) ′k dt
Z0 1
t3i + t5 j + t4k · 1i + 2tj + 3t2k dt
=
Z0 1
=
(t3 + 2t6 + 3t6)dt
0
4
1
t
5t7
=
+
4
7 0
27
= .
28
1.4.2 Path independence and conservative vector fields
Definition 1.46. If C is a parametric curve in the Cartesian plane of the form
r(t) = f (t)i + g(t)j
where
a 6 t 6 b,
then the initial point of C is given by the position vector r(a) and the terminal point of C is
given by the position vector r(b). Similarly, one can define the initial and terminal points of a parametric curve
r(t) = f (t)i + g(t)j + h(t)k
a6t6b
in three dimensional space.
Example 1.47. Let F be the vector field
F (x, y) = y 2 i + xj
and let C1 and C2 be the parametric curves defined as
C1: r(t) = (5t − 5)i + (5t − 3)j
C2: r(t) = (4 − t2)i + tj
06t61
− 3 6 t 6 2.
a) Sketch the curves C1 and C2 and verify that C1, C2 have the same initial and terminal point
b) By evaluating the respective line integrals, show that
Z
Z
F · dr F · dr
Answer:
C1
.
C2
a) In the case of C1, we have
x = 5t − 5
.
y = 5t − 3
Note that both x, y are linear functions of t and that the parameter t takes values in a finite
interval 0 6 t 6 1. It follows that C1 is a line segment with initial point r(0) = ( − 5, − 3) and
terminal point r(1) = (0, 2).
In the case of C2, the initial point is r( − 3) = ( − 5, − 3) and terminal point r(2) = (0, 2). To
determine the shape of C2, we eliminate t from
to obtain
x = 4 − t2
y =t
x = 4 − y 2.
26
Vector Calculus
It follows that C2 is a segment of the parabola x = 4 − y 2 that has initial point ( − 5, − 3) and
terminal point (0, 2).
Figure 1.5. Sketch of curves C1 and C2
b) Using the definition of the line integral, we have
Z
Z b
F · dr =
F (x(t), y(t)) · (x ′(t)i + y ′(t)j ) dt
C1
a
Z 1
=
F ((5t − 5), (5t − 3)) · ((5t − 5) ′i + (5t − 3) ′ j ) dt
0
Z 1
(5t − 3)2i + (5t − 5)j · (5i + 5j ) dt
=
Z0 1
=
5(5t − 3)2 + 5(5t − 5) dt
0
Z 1
25t2 − 25t + 4 dt
=5
0
=−
and
Z
C2
5
6
F · dr =
=
=
=
=
Z
Z
b
F (x(t), y(t)) · (x ′(t)i + y ′(t)j ) dt
a
2
Z−3
2
F ((4 − t2), t) · (4 − t2) ′i + (t) ′ j dt
Z−3
2
Z−32
−3
=
and clearly
245
6
Z
C1
t2i + (4 − t2)j · (( − 2t)i + j ) dt
( − 2t)t2 + 4 − t2 dt
− 2t3 − t2 + 4 dt
F · dr Z
C2
F · dr .
Definition 1.48. A curve C is called a piecewise smooth curve if it is a finite union
C = C1 ∪ C2 ∪ ∪ Cn
27
1.4 Line integrals and double integrals
of smooth curves C1, C2, , Cn such that the terminal point of Ci is the initial point of Ci+1.
Figure 1.6. A piecewise smooth C = C1 ∪ C2 ∪ C3
A line integral along a piecewise smooth curve is obtained by adding the line integrals of its individual pieces.
Definition 1.49. Let C = C1 ∪ C2 ∪ ∪ Cn be a piecewise smooth curve. Then the line integral of
the vector field F along the curve C is
Z
Z
Z
Z
F · dr +
F · dr + +
F · dr
F · dr =
C
C1
C2
Cn
Definition 1.50. A line integral
is said to be independent of path if Z
C
Z
C
F · dr
F · dr =
Z
C′
F · dr
for any piecewise smooth curve C ′ which has the same initial point and terminal point as the curve
C.
Example 1.51. Consider curves C1, C2 and the line integral
Z
F · dr
C1
as defined in Example 1.47. This line integral is not independent of path as C2 has the same initial
and terminal points as C1 but
Z
Z
F · dr F · dr .
C1
C2
Recall from Remark 1.25 that the gradient grad(f ) of a function f (x, y) is in fact a vector field.
Definition 1.52. Let F be a vector field defined on Rn1.2. Then F is said to be a conservative
vector field if there exists a function f such that
F = grad(f ).
For such a case, f is called a potential function for the vector field F.
1.3
1.2. In this class we consider only the cases of n = 2 (the Cartesian plane) and n = 3 (three dimensional space)
1.3. A different definition F = − grad(f ) is used in physics; with this alternative definition, the function f gives a more
accurate physical interpretation of the work done by an outside force in moving against the vector field. In this class, we use
the definition F = grad(f ).
28
Vector Calculus
The above is the standard definition of a conservative vector field; however it is not a practical
method of determining whether or not a vector field is conservative. A more useful criterion for
vector fields in R2 is the following
Theorem 1.53. Let
F (x, y) = P (x, y)i + Q(x, y)j.
be a vector field defined on R2 where the component functions P and Q have continuous first partial
derivatives. Then F is conservative if and only if
∂P ∂Q
=
.
∂x
∂y
Example 1.54. Show that the vector field
F (x, y) = (6x + 5y)i + (5x + 4y)j
is conservative and determine a potential function for F (x, y).
Answer: In this case
P (x, y) = 6x + 5y
Q(x, y) = 5x + 4y
and
∂Q
∂P
=5=
∂x
∂y
so it follows that F is conservative. Let f be a potential function for F . Then
grad(f ) = F
⇒
∂f
∂f
i + j = (6x + 5y)i + (5x + 4y)j
∂x
∂y
∂f
= 6x + 5y
∂x
⇒
.
∂f
= 5x + 4y
∂y
(1.2)
Using partial integration1.4 to integrate the first equation of (1.2) with respect to x, we have
f (x, y) = 3x2 + 5xy + C(y).
(1.3)
The function C(y) can be determined by differentiating equation (1.3) with respect to y and using
the second equation of (1.2)
∂f
= 5x + C ′(y) = 5x + 4y
∂y
⇒ C ′(y) = 4y
⇒ C(y) = 2y 2 + K
where K is a constant of integration. We therefore have
f (x, y) = 3x2 + 5xy + 2y 2 + K
and by choosing K = 0 we have that f = 3x2 + 5xy + 2y2 is a potential function for the vector field
F.
The following result specifies exactly what conditions are required for a line integral in Rn to be
independent of path.
1.4. see Example 1.59 on page 31
29
1.4 Line integrals and double integrals
Theorem 1.55. Let F be a vector field on Rn. Then the line integral
Z
F · dr
C
is independent of path if and only if F is a conservative vector field.
Furthermore, the value of a line integral that is independent of path can be determined from the
endpoints of the curve and a potential function of the vector field
Theorem
R 1.56. (Fundamental Theorem of Line Integrals) Let F be a vector field defined on
Rn and C F · dr be a line integral that is independent of path. Then
Z
F · dr = f (r2) − f (r1)
C
where f is a potential function for the the vector field F and r1, r2 are respectively the initial and
terminal points of the curve C.
Example 1.57. Let F be the vector field
F (x, y, z) = x3 y 4 i + x4 y3 j
and let C be the parametric curve defined as
√
C: r(t) = t i + (1 + t3)j
0 6 t 6 1.
a) Show that F is a conservative vector field
b) Find a potential function for F
c) Use the potential function of part (b) to evaluate the line integral
Answer:
R
C
F · dr.
a) The given vector field is of the form F = P (x, y)i + Q(x, y)j where
P (x, y) = x3 y 4
.
Q(x, y) = x4 y 3
As
so it follows that F is conservative.
∂P
∂Q
= 4x3 y 3 =
∂y
∂x
b) Let f be a potential function for F . Then
grad(f ) = F
⇒
∂f
∂f
i + j = x3 y 4 i + x4 y 3 j
∂y
∂x
∂f
= x3 y 4
∂x
⇒
.
∂f
= x4 y 3
∂y
(1.4)
By partially integrating the first equation of (1.4) with respect to x, we have
f (x, y) =
x4 y 4
+ C(y).
4
(1.5)
The function C(y) can be determined by differentiating equation (1.5) with respect to y and
using the second equation of (1.2)
∂f
= x4 y 3 + C ′(y) = x4 y 3
∂y
⇒ C ′(y) = 0
⇒ C(y) = K
30
Vector Calculus
where K is a constant of integration. We have
f (x, y) =
and by choosing K = 0 we have that f =
x4 y 4
4
x4 y 4
+K
4
is a potential function for the vector field F .
c) Using the parametric definition of C, we have that the initial and terminal points of C are
r1 = r(0) = (0, 1)
r2 = r(1) = (1, 2)
and from Theorem 1.56
Z
C
F · dr = f (r2) − f (r1)
= f (1, 2) − f (0, 1)
1 4 2 4 0 41 4
−
4
4
= 4.
=
1.4.3 Double integrals
We give the definition of a double integral over a rectangular region R and then state a theorem that
gives a procedure for evaluating double integrals over a rectangular region. We then note that a
double integral may be interpreted as a volume. Finally we explain the evaluation of double integrals over Type I , Type II and circular regions.
Definition 1.58. Let R be a rectangular region in the Cartesian plane defined by
R = {(x, y)| a 6 x 6 b, c 6 y 6 d}.
Divide the rectangular region R into subrectangles by partitioning the intervals a 6 x 6 b and c 6 y 6
d:
a = x0 < x1 < < xm = b
c = y0 < y1 < < yn = d
and define the subrectangle Rij as
Rij = {(x, y)|xi−1 6 x 6 xi , yj −1 6 y 6 yj }
31
1.4 Line integrals and double integrals
∗
∗
Choose points
(xij , yij ) in each subrectangle Rij . Let ∆Aij denote the area of the subrectangle Rij
and let P denote the length of the largest
diagonal of all subrectangles (note that as we take
smaller subrectangles of R we have that P → 0). Then the double integral of the function f (x,
y) over the rectangular region R is defined to be the limit
ZZ
m X
n
X
∗
∗
f (x, y)dA = lim
f (xij
, yij
)∆Aij
kP k→0
R
if this limit exists.
i=1 j =1
The above definition is not usually used to evaluate double integrals. We shall give a method of evaluating double integrals based on the following procedure of partial integration.
Example 1.59. (of partial integration)
2 y=2
Z 2
xy
xy dy =
2 y=1
1
x22 x12
=
−
2
2
3x
= .
2
Notice in the above procedure of partial integration we
•
integrate a function of two variables with respect to y
•
treat x as a constant when integrating
•
the answer
Z
2
xy dy =
1
is a function of x.
3x
2
We can also integrate partially with respect to x:
Example 1.60.
Z
4
sin(xy) x=4
y
x=3
sin(4 y) sin(3 y)
−
=
y
y
cos(xy) dy =
3
and notice in this case our answer is a function of y.
Using the procedure of partial integration and the following theorem we can evaluate double integrals over rectangular regions.
Theorem 1.61. (Fubini’s Theorem) If the function f (x, y) is continuous at each point in a rectangular region
R = {(x, y)| a 6 x 6 b, c 6 y 6 d}
then
!
!
Z d Z b
Z b Z d
ZZ
f (x, y)dx dy.
(1.6)
f (x, y)dy dx =
f (x, y)dA =
R
a
c
c
a
Note that the integrals within the brackets of (1.6) are evaluated by partial integration. Also notice
that (1.6) implies that the double integral over a rectangular region can be obtained by integrating
with respect to y and then x or integrating with respect to x and then y.
Example 1.62. Evaluate the double integral
ZZ
R
x2 y dA
32
Vector Calculus
where R is the rectangular region
R = {(x, y)| 0 6 x 6 3, 1 6 y 6 2}.
Answer: By Fubini’s Theorem
ZZ
3
2
x2 ydy dx
1
0
Z 3 2 2 y=2
x y
=
dx
2 y=1
0
Z 3 2 2
x 2
x2 12
dx
=
−
2
2
Z0 3 2
3x
=
dx
2
0 3 x=3
x
=
2 x=0
27
= .
2
x2 y dA =
R
Z
Z
Notation 1.63.
i. The integral
is usually denoted as
Z
b
Z
Z
b
a
d
d
f (x, y)dydx.
c
Z
b
Z
d
c
Z
!
f (x, y)dx dy
a
c
is usually denoted as
Z
!
f (x, y)dy dx
c
a
ii. Similarly, the integral
d
Z
b
f (x, y)dxdy.
a
Recall that the graph of a function f (x, y) forms a surface in three dimensional space. Given a function f (x, y) > 0 for each point (x, y) in a rectangular domain R, then the double integral
ZZ
f (x, y)dA
R
can be interpreted as the volume between the graph of f (x, y) and the rectangle R lying in the xy −
plane
33
1.4 Line integrals and double integrals
We now consider double integrals over regions in the Cartesian plane that are not rectangular.
Definition 1.64.
i. A Type I region is a region in the Cartesian plane that may be described as
R = {(x, y)| a 6 x 6 b, f (x) 6 y 6 g(x)}
where f (x) and g(x) are continuous functions of x.
ii. A Type II region is a region in the Cartesian plane that may be described as
R = {(x, y)| k(y) 6 x 6 h(y), c 6 y 6 d}
where k(y) and h(y) are continuous functions of y.
Figure 1.7.
As in the evaluation of a double integral over a rectangular region, evaluating double integrals over
Type I and Type II regions requires the use of partial integration. Note that for a Type I region,
the integration is done with respect to the y variable first and then with respect to x variable. For a
Type II region, the integration is done with respect to the y variable first and then with respect to x
variable.
Theorem 1.65.
i. If the function f (x, y) is continuous at each point in a Type I region
R = {(x, y)| a 6 x 6 b, g(x) 6 y 6 f (x)}
then
ZZ
f (x, y)dA =
R
Z
b
a
Z
f (x)
!
f (x, y)dy dx
g(x)
ii. If the function f (x, y) is continuous at each point in a Type II region
R = {(x, y)| k(y) 6 x 6 h(y), c 6 y 6 d}
then
ZZ
R
f (x, y)dA =
Z
c
d
Z
h(y)
!
f (x, y)dx dy.
k(y)
34
Vector Calculus
Example 1.66. Evaluate the double integral
ZZ
where R = {(x, y)| 0 6 x 6 1, 0 6 y 6
p
x dA
R
1 − x2 }.
Answer: Notice that R is a Type I region. Then
ZZ
R
xdA =
Z
1
0
Z
√
!
1−x2
x dy dx
0
Z
1
√
2
y= 1−x
dx
[xy] y=0
0
Z 1 p
=
x 1 − x2 − x(0) dx
Z0 1 p
=
x 1 − x2 dx
0
#x=1
"
3
(1 − x2) 2
= −
3
x=0
1
= .
3
=
Example 1.67. Evaluate the double integral
ZZ
xy dA
R
y2
where R is the region bounded by the line x = y + 1 and the parabola x = 2 − 3.
Answer: By solving the equations
x= y +1
y2
x= −3
2
⇒ y+1=
y2
−3
2
y2
⇒
y = − 2, 4
and we see that the line x = y + 1 and the parabola x = 2 − 3 intersect at the points ( − 1, − 2) and
(5, 4):
35
1.4 Line integrals and double integrals
From the above diagram we can write R as
R = {(x, y)|
y2
− 3 6 x 6 y + 1, − 2 6 y 6 4}
2
and so the region R is of Type II . Then
ZZ
R
xy dA =
Z
=
4
−2
Z
1
=
2


4
−2
Z
Z

y+1
y2
−3
2
x2 y
2
xy dx dy
x=y+1
x=
y2
−3
2
4
dy
2
(y + 1) y −
−2
Z 4
y2
−3
2
2 !
y dy
1
y5
y 3 + 2y 2 + y − + 3y 3 − 9y dy
2 −2
4
Z 1 4
y5
3
2
=
− + 4y + 2y − 8y dy
2 −2
4
= 36.
=
Some regions in the Cartesian plane are more easily described by the use of polar coordinates.
Definition 1.68. A polar rectangle is a region in the Cartesian plane that may be described in
polar coordinates (r, θ) as
R = {(r, θ)| θ1 6 θ 6 θ2, a 6 r 6 b}
where a and b are real constants such that 0 6 a < b.
Figure 1.8.
Example 1.69. Express the following region
as a polar rectangle.
R = {(x, y)| 9 6 x2 + y 2 6 25, x > 0, y > 0}
Answer: The equations
x2 + y2 = 9
x2 + y2 = 25
describe circles of radius 3 and 5 respectively that have the origin as center. Therefore the inequalities
9 6 x2 + y2 6 25, x > 0 and y > 0
36
Vector Calculus
describe points that lie between the circles and in the first quadrant
and so we can write R as a polar rectangle
π
R = {(r, θ)| 0 6 θ 6 , 3 6 r 6 5}.
2
We shall use the following theorem to evaluate double integrals over regions in the Cartesian plane
that are polar rectangles. Notice that the following theorem is a conversion of a double integral in
xy − coordinates to a double integral in polar coordinates.
Theorem 1.70. If the function f (x, y) is continuous at each point in a polar rectangle
R = {(r, θ)| θ1 6 θ 6 θ2, a 6 r 6 b}
then
ZZ
f (x, y)dA =
R
Z
θ2
θ1
Z
b
!
f (r cosθ, r sinθ)rdr dθ.
a
Example 1.71. Evaluate the following double integral
ZZ
1 − (x2 + y 2) dA
R
where R = {(x, y)| 0 6 x2 + y 2 6 1} by converting to polar coordinates.
Answer: The region R is the unit disc
and so we can write R as a polar rectangle
R = {(r, θ)| 0 6 θ 6 2π, 0 6 r 6 1}.
37
1.5 Green’s theorem
From Theorem 1.70, we have
ZZ
Z 2π Z 1
2
2
2
2
2
2
(1 − (r cos θ + r sin θ))rdr dθ
1 − (x + y ) dA =
R
Z 02π Z 01
(1 − r2)rdr dθ
=
Z0 2π Z0 1
3
(r − r )dr dθ
=
0
0
1
Z 2π 2
r4
r
−
dθ
=
4 0
2
0
Z 2π
1
=
dθ
4
0
π
= .
2
1.5 Green’s theorem
Definition 1.72.
i. A parametric curve C is called closed if its terminal point coincides with its initial point.
ii. A simple curve C is a curve that does not intersect itself anywhere between its endpoints.
Figure 1.9. Examples of curves
Convention: A positive orientation of a simple closed curve C refers to a counterclockwise traversal
of C.
Theorem 1.73. (Green’s Theorem) Let C be a positively oriented, piecewise smooth, simple
closed curve in the Cartesian plane. Let D be the region bounded by C. Let F be the vector field
F (x, y) = P (x, y)i + Q(x, y)j
where P (x, y) and Q(x, y) have continuous partial derivatives on an open region that contains D.
Then
Z
ZZ ∂Q ∂P
dA
F · dr =
−
∂x
∂y
C
D
38
Vector Calculus
Figure 1.10. Region D bounded by curve C
Note that Green’s Theorem states that the line integral around a simple closed curve C may be
obtained by evaluating a double integral over the region D enclosed by C.
Example 1.74. Let C = C1 ∪ C2 ∪ C3 be the simple closed curve
enclosing the triangular region D. Use Green’s Theorem to determine the line integral
Z
(x4i + xyj).dr
(1.7)
C
by evaluating an appropriate double integral.
Answer: By Green’s Theorem
Z
C
F · dr =
ZZ D
∂Q ∂P
−
dA
∂y
∂x
we have that the line integral (1.7) is equal to a double integral
Z
ZZ ∂
∂
(x4i + xyj) · dr =
(xy) − (x4) dA
∂y
∂x
C
ZZD
=
(y − 0)dA
Z ZD
=
y dA
(1.8)
D
and this last integral is a double integral of the function f (x, y) = y over the triangular region D.
Notice that D is a Type I region
39
1.6 Surface integrals
that is, we can describe the region D as the set
D = {(x, y)| 0 6 x 6 1, 0 6 y 6 1 − x}
and therefore
ZZ
y dA =
D
Z
0
1
1−x
Z
(1.9)
y dy dx
0
Substituting (1.9) into (1.8) we have
ZZ
Z
4
y dA
(x i + xyj) · dr =
D
C
=
Z
1
Z
0
1−x
y dy dx
0
and notice that this is the point of Green’s theorem – for a simple closed curve C, a line integral
around C is equal to a double integral over the region enclosed by C
Z
(x4i + xyj).dr =
C
Z
1
0
Z
1−x
y dy dx
0
and we can evaluate this double integral by using partial integration
Z 1 Z 1−x
Z
y dy d x
(x4i + xyj).dr =
0
0
C
=
Z
1
0
y2
2
y=1−x
dx
y=0
1
(1 − x)2
dx
2
0
(1 − x)3 1
= −
6
0
1
=
6
=
Z
1.6 Surface integrals
1.6.1 Parametric surfaces
Recall that some curves C in three dimensional space can be described parametrically
C:
r(t) = x(t)i + y(t)j + z(t)k
for example the curve shown below
a 6 t 6 b,
40
Vector Calculus
may be written parametrically as
r(t) = ti + tj + (1 − t)k
0 6 t 6 1.
We now describe surfaces in three dimensional space by the use of vector functions r(u, v) of two
parameters u and v.
Definition 1.75. A parametric surface S in three dimensional space is obtained by specifying x,
y and z to be continuous functions of parameters u and v
x = f (u, v)
y = g(u, v)
z = h(u, v)
where (u, v) ∈ R and R is a region in the u v − plane. The parametric surface S can be written in
vector form
r(u, v) = f (u, v)i + g(u, v)j + h(u, v)k
(u, v) ∈ R.
Example 1.76. Let S be the (truncated) plane defined by y = 1, 0 6 x, z 6 1.
Then we can describe S a parametric surface by
r(u, v) = ui + j + vk
(u, v) ∈ R.
where R is the region {(u, v)|0 6 u, v 6 1}
in the uv − plane. Notice that each point (u, v) of R defines a unique point in S.
Example 1.77. Describe the cylinder S
x2 + y 2 = 4, 0 6 z 6 3.
as a parametric surface.
41
1.6 Surface integrals
Answer: We can use polar coordinates to describe the cylinder S
as the parametric surface
r(θ, z) = 2cosθi + 2 sin θj + zk
(θ, z) ∈ R
where R is the region {(θ, z)|0 6 θ 6 2π, 0 6 z 6 3}
in the θz − plane.
Lemma 1.78. Let S be a parametric surface
r(u, v) = f (u, v)i + g(u, v)j + h(u, v)k
(u, v) ∈ R
that is smooth. Then a normal vector n to S at the point r(u0, v0) is given by
where ru and rv are the vectors
provided that ru × rv 0.
n = ru × rv
ru =
∂f
∂g
∂h
(u0, v0)i + (u0, v0)j + (u0, v0)k
∂u
∂u
∂u
rv =
∂f
∂g
∂h
(u0, v0)i + (u0, v0)j + (u0, v0)k
∂v
∂v
∂v
Example 1.79. Find a normal vector to the cylinder
S: r(θ, z) = 2cosθi + 2 sin θj + zk
(θ, z) ∈ R
(1.10)
where R is the region {(θ, z)|0 6 θ 6 2π, 0 6 z 6 3} at the point (2, 0, 0).
Answer: In this case the parameters are θ and z. Note that the point (2, 0, 0) corresponds to (θ, z) =
(0, 0) as
r(0, 0) = 2cos0i + 2 sin 0j + 0k
= 2i
42
Vector Calculus
The vector functions rθ(θ, z) and rz (θ, z) are obtained by differentiating (1.10) partially with respect
to θ and z respectively
rθ(θ, z) = − 2sinθi + 2 cos θj + 0k
rz (θ, z) = 0i + 0j + 1k
and at (θ, z) = (0, 0)
rθ(0, 0) = 2 j
rz (0, 0) = k
and the normal vector at the point (2, 0, 0) is
n = rθ(0, 0) × rz(0, 0)
= 2j × k
= 2i
1.6.2 Surface integrals
Definition 1.80. Let S be a smooth parametric surface
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k
(u, v) ∈ R
where R is a region in the uv − plane and let f (x, y, z) be a continuous function defined on S. Then
the surface integral of the function f (x, y, z) over the surface S is
ZZ
ZZ
f (x, y, z) dS =
f (x(u, v), y(u, v), z(u, v))|ru × rv | dA
(1.11)
S
R
Note that equation (1.11) defines a surface integral to be equal to a double integral over a region R
in the uv − plane where u and v are the parameters of the surface S.
Example 1.81. Evaluate the surface integral
ZZ
(x2 y + z 2) dS
S
where S is the part of the cylinder x2 + y 2 = 9 that lies between the plane z = 0 and z = 2.
43
1.6 Surface integrals
Answer: Our definition (1.11) of a surface integral requires that the surface S
be given in parametric form:
S: r(θ, z) = 3cosθi + 3 sin θj + zk
(θ, z) ∈ R
(1.12)
where R is the region {(θ, z)|0 6 θ 6 2π, 0 6 z 6 2}. Note in this case we use the parameters r and θ
instead of u and v. The x, y and z components of S are
x = 3cosθ
y = 3 sin θ z = z.
The vectors rθ and rz are
rθ =
∂x
∂y
∂z
i + j + k = − 3sinθi + 3cosθj + 0k
∂θ
∂θ
∂θ
rz =
and we have
∂y
∂z
∂x
i + j + k = 0i + 0j + k
∂z
∂z
∂z
rθ × rz = ( − 3sinθi + 3cosθj) × k
i
j
k = − 3sinθ 3cos θ 0 0
0
1
= 3cos θ i + 3 sinθj
and therefore
|rθ × rz | = |3cos θ i + 3 sinθj |
√
= 9cos2 θ + 9 sin2θ
= 3.
From the definition of a surface integral
ZZ
ZZ
f (x(θ, z), y(θ, z), z(θ, z)) |rθ × rz | dA
(x2 y + z 2) dS =
S
ZZR
=
(3cosθ)2(3sinθ) + z 2 3 dA
ZR
Z
=3
27cos2θ sinθ + z 2 dA
R
This last integral is a double integral of the function f (θ, z) = 27cos2θ sinθ + z 2 over the rectangular
region R
44
Vector Calculus
in the θz − plane. Therefore
ZZ
S
2
ZZ
2
(x y + z ) dS = 3
27cos2θ sinθ + z 2 dA
Z 2πRZ 2
27cos2θ sinθ + z 2 dzdθ
=3
2
Z0 2π 0
z3
2
dθ
=3
27cos θ sinθz +
3 0
Z 02π 8
=3
54cos2θ sinθ +
dθ
3
0
2π
54cos3θ 8θ
(use substitution u = cos θ)
=3 −
+
3
3 0
= 16π
A surface integral of a function f over a surface S can be interpreted as follows. Suppose that a surface S is subdivided into n smaller surfaces Si.
Choose a point (x∗i , yi∗, zi∗) on each smaller surface Si. Let ∆Ai be the area of the small surface Si
and multiply this area by the value f (x∗i , yi∗, zi∗) of the function f at the point (x∗i , yi∗, zi∗). Hence we
have a ‘weighted’ area
f (x∗i , yi∗, zi∗)∆Ai
for each of the n smaller surfaces Si of the entire surface S. Then the surface integral of the function
f over the surface S is approximately the sum of each of the weighted areas for the Si
ZZ
S
f (x, y, z) dS ≃
n
X
f (x∗i , yi∗, zi∗)∆Ai
(1.13)
i=1
and if the number n of smaller surfaces Si gets larger then the approximation (1.13) gets better, so
we have
ZZ
n
X
(1.14)
f (x, y, z) dS = lim
f (x∗i , yi∗, zi∗)∆Ai .
n→∞
S
i=1
Notice that if the function f (x, y, z) = 1 then from (1.14) we have
ZZ
S
1 dS = lim
n→∞
n
X
∆Ai
i=1
and the right side of (1.14) is nothing but the sum of the areas of each of the smaller surfaces Si.
This sum is obviously equal to the surface area of S. Therefore we have the following lemma
45
1.6 Surface integrals
Lemma 1.82. The surface integral of the constant function f = 1 over the surface S is equal to the
surface area of S
ZZ
1 dS = surface area of S.
S
Example 1.83. Find the surface area of the truncated plane S
r(u, v) = ui + j + vk
(u, v) ∈ R.
where R is the region {(u, v)|0 6 u, v 6 1}.
Answer: The surface S is shown in the following diagram
(see Example 1.76). Clearly the surface area of S is equal to 1; let us verify this by evaluating the
surface integral
ZZ
1 dS.
S
The x, y and z components of S are
x=u
y = 1 z = v.
The vectors ru and rv are
ru =
∂x
∂y
∂z
i + j + k = i + 0j + 0k
∂u
∂u
∂u
rv =
∂x
∂y
∂z
i + j + k = 0i + 0j + k
∂v
∂v
∂v
and we have
ru × rv = i × k
i j k
= 1 0 0 0 0 1
=−j
and so
|ru × rv | = | − j | = 1
From the definition of a surface integral
ZZ
1 ds =
S
=
ZZ
Z ZR
R
1|ru × rv | dA
1 dA
46
Vector Calculus
This last integral is a double integral of the function f (u, v) = 1 over the rectangular region R
in the uv − plane. Therefore
ZZ
S
ZZ
1 dS =
1 dA
Z 1
Z 1R
1dudv
=
Z0 1 0
1
=
[u]0 dv
Z0 1
(1 − 0)dv
=
Z0 1
=
1dv
0
=1
which agrees with our expected answer.
Recall that the graph of a function g(x, y) forms a a surface in three dimensional space. Let R be a
region in the xy − plane and let S be the surface formed as the image of the region R in the graph of
g(x, y):
Then the following lemma expresses a surface integral of a function f (x, y, z) over such a surface S
in terms of the function g(x, y).
Lemma 1.84. Let R be a region in the xy − plane and let S be the image of R in the graph of a
function g(x, y). Then the surface integral of the function f (x, y, z) over the surface S is equal to a
double integral over the region R in the xy − plane:
s
2 2
ZZ
ZZ
∂g
∂g
f (x, y, z) dS =
f (x, y, g(x, y)) 1 +
+
dA
∂x
∂y
S
R
Proof. The surface S can be written parametrically as
r(x, y) = xi + yj + g(x, y)k
(x, y) ∈ R
47
1.6 Surface integrals
with parameters x and y. The vectors rx and ry are
rx =
ry =
and therefore
∂
∂
∂
∂g
(x)i + (y)j + (g(x, y))k = i + 0j + k
∂x
∂x
∂x
∂x
∂
∂
∂
∂g
(x)i + (y)j + (g(x, y))k = 0i + j + k
∂y
∂y
∂y
∂y
∂g
∂g
rx × r y = i + k × j + k
∂y
∂x
i j k 1 0 ∂g =
∂x 0 1 ∂g ∂y =−
∂g
∂g
i− j+k
∂y
∂x
hence
∂g
∂g
|rx × r y | = − i − j + k ∂x
∂y
s
2 2
∂g
∂g
= 1+
+
∂x
∂y
and from the parametric definition of a surface integral in Definition 1.80 we have
ZZ
ZZ
f (x, y, g(x, y))|rx × r y | dA
f (x, y, z) dS =
R
S
s
2 2
ZZ
∂g
∂g
dA
+
f (x, y, g(x, y)) 1 +
=
∂y
∂x
R
which is the desired result.
Example 1.85. Find the surface area of the hemisphere of radius a
p
z = a2 − (x2 + y2) .
Answer: Denote the hemisphere as S
Notice that the hemisphere S is the image of the region R
R: x2 + y 2 6 a2
in the graph of the function
z = g(x, y) =
p
a2 − (x2 + y2) .
48
Vector Calculus
Recall from Lemma 1.82 that the surface area of S is given by the surface integral of the constant
function f = 1 over S, that is
ZZ
1 d Ss.
surface area of S =
S
Now because the hemisphere S is the image of a region R in the graph of the function g(x, y), from
Lemma 1.84 we have
2 2
ZZ s
ZZ
∂g
∂g
dA
+
1+
1dS =
∂y
∂x
R
S
and therefore we have
ZZ
1 dS
surface area of S =
S s
2 2
ZZ
∂g
∂g
1+
=
+
dA
∂x
∂y
R
v
!2
!2
ZZ u
u
y
x
t
dA
+ p
1+ p
=
a2 − (x2 + y2)
a2 − (x2 + y 2)
R
ZZ s
x2 + y 2
1+ 2
=
dA
a − (x2 + y 2)
R
s
ZZ
a2
dA
=
a2 − (x2 + y 2)
R
ZZ
− 1
=a
a2 − (x2 + y 2) 2 dA
R
2
2
where R: x + y 6 a
2
is the disc of radius a
and so we convert to polar coordinates
surface area of S = a
ZZ
a2 − (x2 + y 2)
R
Z
2π
Z
a
− 1
2
dA
a2 − (r2cos2θ + r2sin2θ)
0
0
Z 2π Z a
1
2
2 −2
a −r
=a
rdr dθ
Z 02π 0
r=a
1
− a2 − r 2 2
=a
dθ
r=0
Z0 2π
=a
a dθ
=a
− 1
2
rdrdθ
0
= 2πa2
1.6.3 Surface integrals over vector fields
Definition 1.86. A surface S is called an orientable surface if there exists a continuous function
n on S that assigns a unit normal vector to each point of S. Such a function n is called an orientation of S. There are two possible orientations for any orientable surface.
49
1.6 Surface integrals
An oriented surface S is a surface together with one of the two possible choices of orientation.
Example 1.87. Let S be the cylinder
S: r(θ, z) = 2cosθi + 2 sin θj + zk
(θ, z) ∈ R
where R is the region {(θ, z)|0 6 θ 6 2π, 0 6 z 6 3} defined in Example 1.79. Recall from that
example, the vector
rθ(θ, z) × rz (θ, z) = ( − 2sinθi + 2 cos θj) × k
i
j
k = − 2sinθ 2 cos θ 0 0
0
1
= 2cos θ i + 2 sinθj
is a normal vector at any point r(θ, z) on S. The function n(θ, z) given by
rθ × rz
|rθ × rz |
2cos θ i + 2 sinθj
=√
4cos2 θ + 4 sin2θ
= cos θ i + sinθj
n(θ, z)=
assigns the unit normal vector cos θ i + sinθj to each point on the cylinder S. Therefore
n(θ, z) = cos θ i + sinθj
is an example of an orientation on the surface S. Notice that
n1(θ, z) = − n(θ, z) = − (cos θ i + sinθj)
is the other possible orientation of the cylinder S.
We now define the surface integral over a vector field.
Definition 1.88. If F (x, y, z) is a continuous vector field defined on a smooth parametric surface S
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k
(u, v) ∈ R
where R is a region in the uv − plane. Let S have an orientation n given by
n=
ru × rv
.
|ru × rv |
Then the surface integral of the vector field F (x, y, z) over the surface S is
ZZ
ZZ
F (x(u, v), y(u, v), z(u, v)) · (ru × rv) dA.
F .n dS =
S
(1.15)
R
Note that equation (1.15) defines a surface integral of a vector field over a surface to be equal to a
double integral over a region R in the uv − plane where u and v are the parameters of the surface S.
50
Vector Calculus
Example 1.89. Evaluate the surface integral
ZZ
S
where S is the truncated plane1.5
(2xi + 3xzj + y 2k) · n dS
r(u, v) = ui + j + vk
(u, v) ∈ R
and R is the region {(u, v)|0 6 u, v 6 1}.
Answer: The x, y and z components of S are
x=u
y = 1 z = v.
The vectors ru and rv are
ru =
∂x
∂y
∂z
i + j + k = i + 0j + 0k
∂u
∂u
∂u
rv =
∂x
∂y
∂z
i + j + k = 0i + 0j + k
∂v
∂v
∂v
and we have
ru × rv = i × k
i j k
= 1 0 0 0 0 1
=−j
In this case the vector field F is
F (x, y, z) = 2xi + 3xzj + y 2k
From the definition of a surface integral of a vector field over a surface
ZZ
S
ZZ
F (x(u, v), y(u, v), z(u, v)).(ru × rv) dA
(2xi + 3xzj + y2k).n dS =
Z ZR
(2ui + 3uvj + 12k).( − j) dA
=
R
ZZ
− 3uv dA
=
R
This last integral is a double integral of the function f (u, v) = − 3uv over the rectangular region R
1.5. See Example 1.76
51
1.6 Surface integrals
in the uv − plane. Therefore
ZZ
ZZ
− 3uv dA
(2xi + 3xzj + y 2k).n dS =
S
Z 1
Z 1R
− 3uv dudv
=
1
Z0 1 0
− 3u2v
=
dv
2
0
Z 01 − 3(1)2v − 3(0)2v
=
−
dv
2
2
Z0 1
− 3v
dv
=
2
0
−3
=
4
If the vector field F represents the velocity of a fluid moving
R R in three dimensional space, then the
F .n d S can be interpreted as the
surface integral of the vector field F over the surface S
S
volume of fluid flowing through the surface S in unit time. Consider the following example.
Example 1.90. Let F (x, y, z) = y j be a vector field in three dimensional space. Evaluate the following surface integrals
R R
F .n dS where S1 is the parametric surface
i.
S
1
r(u, v) = i + uj + vk
where R is the region {(u, v)|0 6 u, v 6 1}
R R
F .n dS where S2 is the parametric surface
ii.
S2
r(u, v) = vi + j + uk
(u, v) ∈ R.
(u, v) ∈ R.
where R is the region {(u, v)|0 6 u, v 6 1}.
Answer:
i. Note that the vector field F (x, y, z) = y j is in the direction of j at each point (x, y, z) in
three dimensional space. The surface S1 is the truncated plane shown in the diagram below
and
R Rnotice S1 is parallel to the vector field F . If we regard F as the velocity of a fluid and as
F .n d S can be interpreted as the volume of fluid flowing through the surface S1 in
S1
unit time, we expect
ZZ
F .n dS=0
S1
52
Vector Calculus
as there is no fluid flowing through S1 (only along S1). We verify this answer: the x, y and z
components of S1 are
x = 1 y = u z = v.
The vectors ru and rv are
ru =
∂x
∂y
∂z
i+
j + k = 0i + j + 0k
∂u
∂u
∂u
rv =
∂y
∂z
∂x
i + j + k = 0i + 0j + k
∂v
∂v
∂v
and we have
ru × rv = j × k
i j k
= 0 1 0 0 0 1
=i
and by therefore the surface integral of the vector field F over the surface S1 is
ZZ
ZZ
F (x(u, v), y(u, v), z(u, v)).(ru × rv ) dA
F .n dS=
S1
Z ZR
(uj) · i dA
=
Z ZR
0 dA
=
R
=0
ii. In this case the vector field F is perpendicular to the surface S2
and if we again interpret
S2 in unit time, we expect
R R
S2
F .n d S as the volume of fluid flowing through the surface
ZZ
F .n dS > 0.
S2
We verify this answer: the x, y and z components of S2 are
x = v y = 1 z = u.
The vectors ru and rv are
and we have
ru =
∂y
∂z
∂x
i+
j + k = 0i + 0j + k
∂u
∂u
∂u
rv =
∂x
∂y
∂z
i + j + k = i + 0j + 0k
∂v
∂v
∂v
ru × rv = k × i
i j k
= 0 0 1 1 0 0
=j
53
1.6 Surface integrals
and by therefore the surface integral of the vector field F over the surface S1 is
ZZ
ZZ
F (x(u, v), y(u, v), z(u, v)).(ru × rv ) dA
F .n dS=
S2
Z ZR
(1 j) · j dA
=
Z ZR
1dA
=
R
This last integral is a double integral of the function f (u, v) = u over the rectangular region
R
in the uv − plane. Therefore
ZZ
S2
ZZ
F .n dS=
u dA
R
Z 1Z
1
=
1 dudv
Z0 1 0
1
=
[u]0 dv
Z0 1
1dv
=
0
= 1.
Consider a surface S that is a subset of the graph of a function g(x, y), to be specific, let R be a
region in the xy − plane and let S be the oriented surface formed as the image of the region R in the
graph of g(x, y) where we assume that the orientation of S is chosen to point upward
Then the following lemma expresses a surface integral of a vector field F (x, y, z) over such an oriented surface S in terms of the function g(x, y).
54
Vector Calculus
Lemma 1.91. Let R be a region in the xy − plane and let S be the oriented surface formed as the
image of R in the graph of a function g(x, y) where the orientation of S is chosen to point upward.
Then the surface integral of the vector field
F (x, y, z) = P (x, y, z)i + Q(x, y, z)j + T (x, y, z)k
over the oriented surface S is equal to a double integral over the region R in the xy − plane:
ZZ ZZ
∂g
∂g
−P
F .n dS =
− Q + T dA
∂x
∂y
R
S
Example 1.92. Evaluate the surface integral
ZZ
(xi + yj + zk).n dS
S
where S is the part of the paraboloid z = 1 − x2 − y2 that lies above the plane z = 0. Assume S has
an upward orientation.
Answer: Note that the paraboloid z = 1 − x2 − y 2 intersects the plane z = 0 where
1 − x2 − y 2 = 0
that is, the paraboloid intersects the plane z = 0 at the circle x2 + y 2 = 1.
From the diagram, we see that the surface S is the image of the region
R: x2 + y 2 6 1
in the graph of the function g(x, y) = 1 − x2 − y 2. The components of the given vector field are
P (x, y, z) = x Q(z, y, z) = y T (x, y, z) = z
and therefore from Lemma 1.91 the surface integral of the given vector field over the oriented surface
S is
ZZ ZZ
∂g
∂g
− Q + T dA
−P
(xi + yj + zk).n dS =
∂y
∂x
R
S
ZZ
=
( − x( − 2x) − y( − 2y) + z)dA
Z ZR
=
− x( − 2x) − y( − 2y) + 1 − x2 − y 2 dA
R
55
1.6 Surface integrals
because z = 1 − x2 − y2. Hence
ZZ
(xi + yj + zk).n dS =
ZZ
R
S
where the region R is the unit disc
and so we convert to polar coordinates
ZZ
ZZ
(xi + yj + zk).n dS =
S
Z
R
2π
Z
1 + x2 + y 2 dA
1 + x2 + y2 dA
1
1 + r 2cos2θ + r2sin2θ rdrdθ
Z0 2π 0 Z 1
2
(1 + r )rdr dθ
=
0
0
1 !
Z 2π 2
r4
r
+
=
dθ
2
4 0
0
Z 2π
3
dθ
=
4
0
3π
= .
2
=
56
Vector Calculus
1.7 Triple integrals and Divergence theorem
1.7.1 Triple integrals
Definition 1.93. Let B be a rectangular box in three dimension defined by
B = {(x, y, z)| a 6 x 6 b, c 6 y 6 d, r 6 z 6 s}.
Divide the rectangular box B into suboxes by partitioning the intervals a 6 x 6 b , c 6 y 6 d and r 6
z 6 s:
a = x0 < x1 < < xm = b
c = y0 < y1 < < yn = d
r = z 0 < z1 < < zl = s
and define the subbox Bijk as
Bijk = {(x, y, z)|xi−1 6 x 6 xi , yj −1 6 y 6 yj , zk−1 6 y 6 zk }
∗
∗
∗
Choose points (xijk
, yijk
, zijk
) in each subbox Bijk . Let ∆Vijk denote the volume of the subrectangle
Bijk and let P denote the length of the longest diagonal of all subboxes (note that as we take
smaller subboxes of B we have that P → 0). Then the triple integral of the function f (x, y, z)
over the rectangular box B is defined to be the limit
ZZZ
m X
n X
l
X
∗
∗
∗
f (x, y, z)dV = lim
f (xijk
, yijk
, zijk
)∆Vijk
B
kP k→0
i=1 j =1 k=1
if this limit exists.
The above definition is not usually used to evaluate triple integrals. The method of evaluating triple
integrals over rectangular boxes is based on the following theorem.
Theorem 1.94. (Fubini’s Theorem for triple integrals) If the function f (x, y, z) is continuous
at each point in a rectangular box
B = {(x, y)| a 6 x 6 b, c 6 y 6 d, r 6 z 6 s}
then
ZZZ
B
f (x, y, z)dV =
Z
a
b
Z
c
d
Z
s
f (x, y, z)dz dy dx.
(1.16)
r
From the above theorem we see that the evaluation of a triple integral requires two instances of partial integration.
57
1.7 Triple integrals and Divergence theorem
Example 1.95. Evaluate the triple integral
ZZZ
where B is the rectangular box
xyz 2 dV
B
B = {(x, y, z)| 0 6 x 6 1, − 1 6 y 6 2, 0 6 z 6 3}.
Answer: By Fubini’s Theorem for triple integrals
Z 1Z 2 Z 3
ZZZ
xyz 2 dz dy dx
xyz 2 dV =
B
0Z 3
Z 01 Z −1
2
2
xyz dz dy dx
=
0
0
−1
z=3
Z 1Z 2
xyz 3
=
dy dx
3 z=0
Z0 1 Z−1
2
9xy dy dx
=
−1
0
Z 1
9xy 2 y=2
=
dx
2
0
y=−1
Z 1
27x
dx
=
2
0 2 1
27x
=
4
0
27
=
4
We also wish to evaluate triple integrals over cylindrical solids that are centered around the z − axis.
The following diagram illustrates the solid
V = {(x, y, z)| x2 + y 2 6 a2, m 6 z 6 l }
that is bounded by the cylinder x2 + y 2 6 a2 and the horizontal planes z = l, z = m.
If we use the tranformation
x = r cosθ
y = r sin θ
z=z
then the set V can be rewritten in cylindrical coordinates (r, θ, z)
V = {(r, θ, z)|0 6 r 6 a, 0 6 θ 6 2π, m 6 z 6 l}.
58
Vector Calculus
Theorem 1.96. If the function f (x, y, z) is continuous at each point in a cylindrical solid
V = {(x, y, z)| x2 + y 2 6 a2, m 6 z 6 l }
then by converting to cylindrical coordinates
Z 2π Z
ZZZ
f (x, y, z)dV =
0
V
a
0
Z
l
f (r cosθ, r sinθ, z)rdz dr dθ.
m
1.7.2 Divergence theorem
Theorem 1.97. (Divergence theorem) Let E be a solid region whose boundary surface S has a
outward orientation. Let F be a vector field whose component functions have continuous partial
derivatives on an open region that contains E. Then
ZZZ
ZZ
div(F )dV
F .n dS =
E
S
Note that the Divergence theorem states that a surface integral over a surface S that encloses a solid
region E may be obtained by evaluating a triple integral the solid E. We illustrate the Divergence
theorem by the following examples.
Example 1.98. Verify the Divergence theorem for the vector field F
F (x, y, z) = 3xi + xyj + 2xzk
and the solid region E that is the cube bounded by the six planes
x=0 y=0 z=0
x = 1 y = 1 z = 1.
Answer: We evaluate the surface integral
ZZ
(3xi + xyj + 2xzk).n ds
and the triple integral
S
ZZZ
div(3xi + xyj + 2xzk)dV .
E
and verify that these are equal. The solid cube E
has boundary surface S consisting of the six truncated planes S1, S2, S3, S4, S5 and S6
59
1.7 Triple integrals and Divergence theorem
Figure 1.11.
and each of these six planes has an orientation that points out of the solid E.
We first evaluate the triple integral
ZZZ ZZZ
∂
∂
∂
div(3xi + xyj + 2xzk)dV =
(3x) + (xy) + (2xz) dV
∂x
∂y
∂z
E
E
ZZZ
=
(3 + x + 2x) dV
E
Z
1
Z
1
Z
1
(by Theorem 1.94)
(3 + 3x) dzdydx
Z 01 Z 01 0
z=1
=
[3z + 3xz]z=0 dydx
Z0 1 Z0 1
(3 + 3x)dydx
=
Z0 1 0
y=1
=
[3y + 3xy] y=0 dx
Z 01
=
(3 + 3x) dx
0
x=1
3x2
= 3x +
2 x=0
9
=
2
=
Next we evaluate the surface integral of F over S. As the surface S consists of S1, S2, S3, S4,
S5 and S6 we have
ZZ
ZZ
ZZ
ZZ
F .n dS
F .n dS +
F .n dS +
F .n dS =
S
Z SZ3
Z SZ2
Z SZ1
(1.17)
F .n dSs
F .n dS +
F .n dS +
+
S4
S5
S6
R R
F .n dS on the right hand side of (1.17) is evaluated by first expressing Si as paraand each
Si
metric surface r(u, v) and then using the definition of a surface integral
ZZ
ZZ
F (x(u, v), y(u, v), z(u, v)).(ru × rv ) dA.
(1.18)
F .n dS=
Si
R
Note that we should choose each parametrization r(u, v) of Si such that the normal vector ru × rv
has the same direction with the outward normal vector of Si as shown in Figure 1.11
R (if
R the normal
F .n ds on
vector ru × rv is in the opposite direction of the outward normal vector of Si then
Si
the right hand side of (1.17) should be adjusted by a minus sign).
60
Vector Calculus
S1
A parametrization for S1 is
r(u, v) = vi + j + uk
(u, v) ∈ R
where R is the rectangular region
Figure 1.12.
The vectors ru and rv are
and
ru =
∂x
∂y
∂z
i + j + k = 0i + 0j + k
∂u
∂u
∂u
rv =
∂x
∂y
∂z
i + j + k = i + 0j + 0k
∂v
∂v
∂v
ru × rv = k × i
i j k
= 0 0 1 1 0 0
= j.
Notice that the normal vector ru × rv = j points in the the same direction with the outward normal
vector of S1 as shown in Figure 1.11. Using (1.18)
ZZ
ZZ
(3vi + vj + 2uvk).(ru × rv) dA
(3xi + xyj + 2xzk).n dS =
S1
ZZR
=
(3vi + vj + 2uvk).(j) dA
Z ZR
v dA
=
Z 1 RZ 1
v dv du (by Theorem 1.61)
=
0
0
Z 1 2 1
v
=
du
2 0
Z 01
1
du
=
0 2
1
=
2
and therefore
ZZ
F .n dS =
S1
1
2
61
1.7 Triple integrals and Divergence theorem
S2
A parametrization for S2 is
r(u, v) = ui + 0j + vk
(u, v) ∈ R
(1.19)
where R is the rectangular region shown in Figure 1.12. Notice the difference with the parametrization for S1 – the i and k components are exchanged. This causes the normal vector ru × rv to point
in the opposite direction
∂y
∂z
∂x
ru = i + j + k = i + 0j + 0k
∂u
∂u
∂u
rv =
∂x
∂y
∂z
i + j + k = 0i + 0j + k
∂v
∂v
∂v
and
ru × rv = i × k
i j k
= 1 0 0 0 0 1
= − j.
The parametrization (1.19) is chosen so that the normal vector ru × rv = − j points in the the same
direction with the outward normal vector of S2 as shown in Figure 1.11. Using (1.18)
ZZ
S2
ZZ
(3ui + 0j + 2uvk).(ru × rv) dA
(3xi + xyj + 2xzk).n dS =
Z ZR
(3ui + 0j + 2uvk).( − j) dA
=
Z ZR
0 dA
=
Z 1 RZ 1
=
0 dv du
0
0
=0
and therefore
ZZ
F .n dS = 0
S2
S3
A parametrization for S3 is
r(u, v) = i + uj + vk
(u, v) ∈ R
where R is the rectangular region shown in Figure 1.12. The vectors ru and rv are
ru =
∂x
∂y
∂z
i + j + k = 0i + j + 0k
∂u
∂u
∂u
rv =
∂x
∂y
∂z
i + j + k = 0i + 0j + k
∂v
∂v
∂v
and
ru × rv = j × k
i j k
= 0 1 0 0 0 1
= i.
62
Vector Calculus
The surface integral is
ZZ
S3
ZZ
(3i + uj + 2vk).(ru × rv) dA
(3xi + xyj + 2xzk).n dS =
Z ZR
(3i + uj + 2vk).(i) dA
=
Z ZR
3 dA
=
Z 1 RZ 1
3 dv du
=
0
0
=3
and therefore
ZZ
F .n dS = 3
S3
S4
A parametrization for S4 is
r(u, v) = 0i + vj + uk
(u, v) ∈ R
where R is the rectangular region shown in Figure 1.12. The vectors ru and rv are
ru =
∂y
∂z
∂x
i + j + k = 0i + 0j + k
∂u
∂u
∂u
rv =
∂x
∂y
∂z
i + j + k = 0i + j + 0k
∂v
∂v
∂v
and
ru × rv = i × k
i j k
= 0 0 1 0 1 0
= − i.
The surface integral is
ZZ
S4
ZZ
(0i + 0j + 0k).(ru × rv) dA
(3xi + xyj + 2xzk).n dS =
Z ZR
(0i + 0j + 0k).( − i) dA
=
Z ZR
0 dA
=
Z 1 RZ 1
0 dv du
=
0
0
=0
and therefore
ZZ
F .n dS = 0
S4
S5
A parametrization for S5 is
r(u, v) = ui + vj + 1k
(u, v) ∈ R
where R is the rectangular region shown in Figure 1.12. The vectors ru and rv are
ru =
∂y
∂z
∂x
i + j + k = i + 0j + 0k
∂u
∂u
∂u
rv =
∂x
∂y
∂z
i + j + k = 0i + j + 0k
∂v
∂v
∂v
63
1.7 Triple integrals and Divergence theorem
and
ru × rv = i × j
i j k
= 1 0 0 0 1 0
= k.
The surface integral is
ZZ
S5
ZZ
(3ui + uvj + 2uk).(ru × rv) dA
(3xi + xyj + 2xzk).n dS =
ZZR
(3ui + uvj + 2uk).(k) dA
=
Z ZR
2u dA
=
Z 1 RZ 1
2u dv du
=
0
0
=1
and therefore
ZZ
F .n dS = 1
S5
S6
A parametrization for S6 is
r(u, v) = vi + uj + 0k
(u, v) ∈ R
where R is the rectangular region shown in Figure 1.12. The vectors ru and rv are
ru =
∂x
∂y
∂z
i + j + k = 0i + j + 0k
∂u
∂u
∂u
rv =
∂y
∂z
∂x
i + j + k = i + 0j + 0k
∂v
∂v
∂v
and
ru × rv = j × i
i j k
= 0 1 0 1 0 0
= − k.
The surface integral is
ZZ
S6
ZZ
(3vi + uvj + 0k).(ru × rv ) dA
(3xi + xyj + 2xzk).n dS =
ZZR
(3vi + uvj + 0k).( − k) dA
=
Z ZR
0 dA
=
Z 1 RZ 1
0 dv du
=
0
0
=0
and therefore
ZZ
S6
F .n dS = 0
64
Vector Calculus
So we have
ZZ
F .n dS =
S
ZZ
+
F .n dS +
Z SZ1
ZZ
F .n dS +
F .n dS +
Z SZ2
ZZ
F .n dS +
F .n dS
S6
S5
S4
F .n dS
Z SZ3
1
+0+3+0+1+0
2
9
= .
2
=
Hence we have
ZZ
F .n dS =
ZZZ
div(F )dV =
E
S
9
2
which verifies the Divergence theorem.
R R
F .n d S
One application of the Divergence theorem is in the evaluation of surface integrals
S
where S is the boundary surface of a solid region E. As we saw in the previous example, the evaluation of such surface
R R R integrals can be tedious. It is sometimes easier to calculate the corresponding
div(F )dV ; by the Divergence theorem
triple integral
E
ZZZ
ZZ
div(F )dV
F .n dS =
E
S
this gives the answer of the required surface integral.
Example 1.99. Use the Divergence Theorem to evaluate the surface integral
ZZ
(yezi + y2 j + cos (xy)k).n dS
S
where S is the surface of the solid bounded by the cylinder x2 + y 2 = 9 and the planes z = 0 and z =
2.
Answer: The closed cylinder S below
encloses the solid region
E = {(x, y, z)|x2 + y 2 ≤ 9 and 0 ≤ z ≤ 2}
From the Divergence Theorem
ZZ
S
F .ndS =
ZZZ
divF dV
E
65
1.7 Triple integrals and Divergence theorem
we have that our required surface integral I can be expressed as a triple integral
ZZZ ZZ
∂
∂
∂
(yez ) + (y 2) + (cos(xy)) dxdy dz
(yez i + y 2 j + cos(xy)k).n dS =
I=
∂y
∂z
∂x
S
Z Z ZE
2y dxdy dz
=
E
From Theorem 1.96 we can convert to cylindrical coordinates
ZZZ
2y dxdy dz
I=
Z 2π ZE 3 Z 2
2r sinθ r dzdrdθ
=
Z 02π Z 03 0
2
z=2
2r sinθz z=0 drdθ
=
Z 02π Z 03
4r2sinθdrdθ
=
0
0
r=3
Z 2π 3
4r sinθ
=
dθ
3
r=0
Z0 2π
=
36 sinθdθ
0
=0
and so the surface integral
ZZ
S
(yez i + y 2 j + cos(xy)k).n dS = 0.
Chapter 2
Laplace transforms
The Laplace transform changes a function y(t) in the t variable to a corresponding function Y (s) in
the new independent variable s.
The Laplace transform provides a method of solving differential and integral equations by converting
these to algebraic expressions as we shall see in the following sections.
2.1 Definition and existence of Laplace transforms
2.1.1 Improper integrals
Recall that a definite integral has two finite limits of integration
Z b
f (t)dt.
a
If we require that one of these limits of integration be ∞ then we need an improper integral –
Definition 2.1. The improper integral
Z
is defined as the limit
Z
a
∞
∞
(2.1)
g(t)dt
a
g(t)dt = lim
R→∞
Z
R
g(t)dt.
(2.2)
a
The improper integral (2.1) is said to converge if the limit of (2.2) exists and is finite. Otherwise
the improper integral is said to diverge.
Example 2.2. Determine if the following improper integrals converge or diverge
R ∞ −t
e dt
i.
1
R ∞
ii. 1 et dt
Answer:
R ∞
R R
i. 1 e −t dt=limR→∞ 1 e−t dt
−t R
e
=limR→∞ − 1
1
=limR→∞ e−1 − e−R
=limR→∞ e−1 − limR→∞ e−R
= e−1 − limR→∞ e−R
(as e−1 is a constant)
−1
=e
(e−R → 0 as R → ∞)
67
68
Laplace transforms
and therefore
R
∞
1
e −t dt converges. Note by defintion of the improper integral
Z ∞
e −t dt =e−1.
1
ii.
R
∞
1
R t
e dt
1
t R
=limR→∞ e 1
=limR→∞ eR − e
=limR→∞ eR − limR→∞
= limR→∞ eR − e
e t dt=limR→∞
R
e
(as e is a constant)
=∞
(eR → ∞ as R → ∞)
R ∞
and therefore 1 e t dt diverges.
R ∞
R ∞ −t
e dt and 1 e t dt respecThe shaded areas of the graphs below represent the integrals of
1
tively. Notice that the shaded area in the both graphs increase as t increases, however the rate at
which the area in the first graph increases is much
smaller than the rate at which the area in the
R ∞
e t dt diverges.
second graph increases. This should suggest that
1
Figure 2.1.
Notice in the above graph that e−t tends to zero as t gets larger, in other words
lim e−t = 0,
t→∞
also notice that et does not tend to zero as t gets larger
lim et = ∞.
t→∞
From Example 2.2 we saw that
seem that if
R
∞
1
e−td t converges and
R
∞
1
etd t diverges. Therefore it would
lim g(t) = ∞
t→∞
then
Z
∞
g(t)dt
a
diverges. This is in fact the case, we state this fact formally:
Lemma 2.3. If limt→∞ g(t) = ∞ then the improper integral
R
∞
a
g(t)dt diverges.
69
2.1 Definition and existence of Laplace transforms
Proof. (Not required). The limt→∞ g(t) is ∞ if, by definition, given any K > 0 there exists a t −
value t0 such that g(t) > K for all t > t0. Fix such a K and t0. Then
!
Z
Z
Z
R
R→∞
R
t0
g(t)dt = lim
lim
g(t)dt +
R→∞
a
=
t0
Z
g(t)dt + lim
R→∞
a
>
=
t0
Z
a
R
Z
g(t)dt + lim
Z a t0
g(t)dt
t0
a
R→∞
g(t)dt
t0
Z
R
because g(t) > K
Kdt
t0
g(t)dt + lim (KR − Kt0)
R→∞
=∞
and hence
R
∞
a
g(t)dt diverges.
2.1.2 Definition and examples of Laplace tranform
Definition 2.4. The Laplace transform of the function y(t), denoted by L[y(t)], is
Z ∞
L[y(t)] =
e −sty(t)dt
(2.3)
0
and is defined for those values of s for which the improper integral (2.3) converges. Note that L[y(t)]
is a function of s and we denote
L[y(t)] = Y (s).
Example 2.5. Use the definition of the Laplace transform to determine L[ekt] where k is a real constant.
kt
L[e ] =
Z
∞
Z0
e −s t ektdt
∞
e −(s−k )t dt
Z R
= lim
e −(s−k )t dt
=
(2.4)
0
R→∞
and when s = k we have
lim
R→∞
Z
R
e
−(s−k )t
0
dt = lim
R→∞
0
Z
R
1dt
0
(2.5)
= lim R
R→∞
=∞
and therefore L[ekt] does not exist for s = k. Now consider when s k
"
#R
Z R
e −(s−k )t
−(s−k )t
e
dt = lim
lim
R→∞ − (s − k)
R→∞ 0
0
= lim
R→∞
=
1
e −(s−k )R
−
s−k
s−k
!
e −(s−k )R
1
− lim
s−k
s−k
R→∞
70
Laplace transforms
From the graphs of Figure 2.1 we see that
lim e−R = 0 and lim eR = ∞
R→∞
and in general
lim
R→∞
R→∞
eaR =
0
∞
if a < 0
if a > 0
(2.6)
that is, limR→∞ e aR depends only on the sign of a. Hence
0
if − (s − k) < 0
−(s−k )R
lim e
=
∞
if − (s − k) > 0
R→∞
and therefore
1
e −(s−k )R
L[ek t] =
− lim
s−k
R→∞ s− k
1
1
lim e −(s−k )R
=
−
s − k R→∞
s−k
(2.7)
(2.8)
and from (2.7) the limit in (2.8) is zero only when s > k. So we have
L[ekt] =
and this holds only when s > k.
1
s−k
(2.9)
Note by setting k = 0 in (2.9) we get
L[1] =
1
s
when s > 0.
(2.10)
Example 2.6. Use the definition of the Laplace transform to show that
L[sin kt] =
when s > 0.
Answer:
Z
k
s2 + k 2
∞
e −st sin ktdt
0
Z R
e −s t sin kt dt.
= lim
L[sin kt] =
R→∞
Using the integration by parts formula
Z
a
with
b
0
v du = [ uv ] ba −
Z
b
u dv
a
du = e−st dt
e−st
dv = k cos kt dt
u=−
s
−st
R
Z R
e sin kt
k
e −st sin kt dt = −
+
e −st cos kt dt
s
s 0
0
v = sin kt
gives
Z
R
0
Integrate
R
R
0
(2.11)
e −st cos kt by parts again with
du = e−st dt
e−st
dv = − k sin kt dt
u=−
s
−st
R
Z R
e cos kt
k
e −st cos kt dt = −
e −st sin kt dt.
−
s
s
0
0
v = cos kt
to get
Z
0
R
(2.12)
71
2.1 Definition and existence of Laplace transforms
Substituting (2.12) into (2.11) gives
−s t
R
R
Z
Z R
k e−s tcos kt
k 2 R −st
e sin kt
−
− 2
e
sin kt dt.
e −st sin kt dt = −
s
s 0
s
s
0
0
0
R R
and solving for 0 e −st sin kt dt in (2.13) gives
s2 + k 2
s2
Z
R
0
e −st sin kt dt = −
R k R
1 −st
e sin kt 0 − 2 e−stcos kt 0
s
s
(2.13)
(2.14)
and evaluating the right side of (2.14) using the limits of integration gives
s2 + k 2
s2
Z
R
0
k
k
1
e −st sin kt dt = − e−sRsin kR − 2 e−s Rcos kR + 2
s
s
s
and taking the limit limR→∞ of both sides of (2.15) we have
!
Z R
k −s R
k
s2 + k 2
1 −sR
−st
lim
e
sin
kR
−
e
cos
kR
+
−
e
sin
kt
dt
=
lim
.
s2
s2
s2 R→∞
s
R→∞
0
(2.15)
(2.16)
Notice as − 1 6 sin kR 6 1 implies that
− e−sR 6 e−sRsin kR 6 e−sR
and taking limits gives
− lim e−sR 6 lim e−sRsin kR 6 lim e−sR
R→∞
R→∞
R→∞
(2.17)
Now from equation (2.6) above if s > 0 (and only if s > 0) then
lim e−sR = 0.
R→∞
So when s > 0 the inequality (2.17) simplifies to
0 6 lim
R→∞
and therefore
e−sRsin kR 6 0
lim e−sRsin kR = 0 .
(2.18)
R→∞
Similarly one can show when s > 0 that
lim e−sRcos kR = 0,
(2.19)
R→∞
and substituting (2.18) and (2.19) into the right side of (2.16) we have
!
Z R
k
s2 + k 2
−st
e
sin kt dt = 2
lim
s
s2 R→∞
0
!
Z R
k
−st
e
sin kt dt = 2
⇒
lim
s + k2
R→∞
0
when s > 0.
⇒
L[sin kt] =
k
s2 + k 2
Example 2.7. Use the definition of the Laplace transform to show that when n = 0, 1, 2, L[tn] =
when s > 0.
n!
sn+1
(2.20)
72
Laplace transforms
Answer: From (2.10) we know that
1
s
0!
0
⇒ L[t ] = 0+1
s
L[1] =
when s > 0
when s > 0
and so (2.20) holds when n = 0. We therefore need to consider when n = 1, 2, 3, Z ∞
L[tn] =
e −st tn dt
0
Z R
e −st tn dt.
= lim
R→∞
0
Using the integration by parts formula with
du = e−st dt
e−st
dv = ntn−1 dt
u=−
s
−st n R
Z
e t
n R −st n−1
e
t
dt.
e −st tn dt = −
+
s
s 0
0
v = tn
gives
Z
R
0
Taking the limit limR→∞ of both sides of (2.21) we have
!
−st n R !
Z R
e t
n
e −st tn dt = lim −
lim
+ lim
s
s R→∞
R→∞
R→∞
0
0
Z
(2.21)
R
e −st tn−1 dt
0
and by using the definition of the Laplace transforms L[tn] and L[tn−1] we have
−st n R !
e t
n
n
L[t ]= lim −
+ L[tn−1].
s
s
R→∞
0
Now
−st n R !
e t
e−sRRn
lim −
= lim −
s
s
R→∞
R→∞
0
n
R
= lim − sR
se
R→∞
!
(2.22)
∞
and this limit is of the indeterminate form ∞ . Applying L’Hospitals rule by differentiating the
numerator and denominator n times with respect to R, we have
n!
Rn
lim − sR = lim − n+1 sR
s e
se
R→∞
R→∞
=0
when s > 0
and therefore
−st n R !
e t
= 0 when s > 0 .
(2.23)
lim −
s
R→∞
0
Substituting (2.23) into (2.22) gives the formula
n
L[tn]= L[tn−1]
s
valid for n = 1, 2, 3, and s > 0.
(2.24)
Notice that by replacing n by n − 1 in (2.24) we have
L[tn−1]=
n − 1 n−2
L[t ]
s
(2.25)
73
2.1 Definition and existence of Laplace transforms
and similarly by replacing n with n − 2, n − 3, , 2, 1 in (2.24) we have, respectively,
n − 2 n−3
L[t ]
s
n − 3 n−4
L[t ]
L[tn−3]=
s
L[tn−2]=
(2.26)
2
L[t2]= L[t1]
s
1 0
L[t]= L[t ]
s
Finally we get (2.20) by using the equations (2.25) and (2.26) to express L[tn] in terms of L[t0] =
1
s
L[1] = :
n
L[tn]= L[tn−1]
s
n n − 1 n−2
L[t ]
=
s
s
n n − 1 n − 2 n−3
=
L[t ]
s
s s
from (2.24)
from (2.25)
=
=
=
=
n n−1 n−2 2 1
s L[t ]
s s
s
n n−1 n−2 2 1 0
s s L[t ]
s
s s
n!
L[t0]
sn
n!
n+1
s
which is our desired result.
2.1.3 Existence of Laplace transform
In Examples 2.5, 2.6 and 2.7 we were able to determine the Laplace transforms of the functions ekt ,
sin kt and tn respectively. We also showed that the Laplace tranforms of those functions existed for
only certain value of s:
y(t)
L[y(t)] Laplace tranform exists when
1
kt
e
s>k
s−k
k
sin kt
s>0
2
s + k2
n!
tn
s>0
sn+1
We now give an example of a function that does not have a Laplace tranform for any value of s.
2
Example 2.8. Show that the Laplace transform of et does not exist for any real value of s.
2
Answer: Recall that from Definition 2.4 L[et ], given by:
Z ∞
2
2
L[et ] =
e −st et dt
(2.27)
0
is defined only for those values of s for which the improper integral on the right converges. Fix any
value of s. If we can show that for this value of s
lim
t→∞
2
e −st et = ∞
74
Laplace transforms
then by Lemma 2.3 the improper integral
Z ∞
0
2
e −st et dt
(2.28)
does not converge for this fixed value of s. And if (2.28) does not converge for any such fixed s then
2
L[et ] does not exist for any s. So we consider (2.27)
lim
t→∞
2
2
e −st et = lim et −st
t→∞
2
and for fixed s, t − st is an increasing quadratic function in t which implies t2 − st goes to infinity
as t → ∞ and therefore
lim
t→∞
which is our desired result.
2
2
e −st et = lim et −st = ∞
(2.29)
t→∞
In Example 2.29 we saw that the Laplace transform of
2
et did not exist because
2
lim et −st = ∞.
(2.30)
t→∞
for any value of s. Equation (2.30) is equivalent to
2
et
lim st = ∞
t→∞ e
2
which indicates that the function et is much larger than est for large values of t as we can see in the
following graph when s = 1.
2
60
et
50
40
30
20
et
10
0
0
0.5
1
1.5
2
2
Example 2.29 indicates that a function such as et that is much larger
than the standard exponential function est as t → ∞ does not have a
Laplace transform.
On the other hand functions that are ‘comparable’ (in the sense that they are less than or equal
to) to exponential functions will have Laplace transforms2.1 as we shall see in Theorem 2.15. The
precise phrase for a function being comparable to an exponential function is that the function is of
exponential order.
Definition 2.9.
A function y(t) is of exponential order if there exist constants A, b and a t − value t0 such that
|y(t)| < Aebt when t > t0.
Example 2.10. Show that y(t) = t3 is of exponential order.
2.1. Provided such functions satisfy certain continuity conditions.
75
2.1 Definition and existence of Laplace transforms
Recall that et has a Taylor series expansion about t = 0, that is
et =
∞
X
tm
m!
m=0
=1+t+
and clearly when t > 0
t2 t3 t4
+ +
+
2
6 24
t2 t3 t4
+
t3 < 6 1 + t + + +
6 24
2
= 6et
and as |t3| = t3 when t > 0 we have
|t3| < 6et when t > 0
and hence t3 is of exponential order where the constants of Definition 2.9 are A = 6, b = 1 and t0 = 0.
We shall need the following definition
Definition 2.11. A function y(t) is piecewise continuous on a finite interval a 6 t 6 b if on this
interval
i. y(t) is defined and takes finite value at each point and
ii. y(t) only has a finite number of discontinuities.
Example 2.12. The function


16 t<2
t
y(t) = t2
26t<3

 12 − t 3 6 t 6 4
is piecewise continuous on 1 6 t 6 4 as we can see from its graph
10
8
6
4
2
0
0
1
2
3
4
that y(t) has one point of discontinuity at t = 2 and is defined and finite everywhere on 1 6 t 6 4.
Example 2.13. Find
R
4
1
y(t)dt of the function y(t) defined in Example 2.12.
The integral of a piecewise defined function is computed by adding the integrals of each of the intervals
Z 4
Z 2
Z 3
Z 4
y(t)dt=
y(t)dt +
y(t)dt +
y(t)dt
1
Z 4 3
Z 3 2
Z 12
(12 − t )dt
t2 dt +
t dt +
=
1
=
2
3 19 17 49
+
+
=
2
3
2
3
3
76
Laplace transforms
To find the Laplace transform of a piecewise defined function y(t) such as the one defined in
Example 2.12, we use the same procedure of adding the integral over each interval as done in
Example 2.13 :
Example 2.14. Sketch the graph of the following function

06 t<1
0
y(t) = t
16t<2

0
t>2
and determine L[y(t)].
Answer: The graph of y(t) is
4
3
2
1
0
1
2
By definition
L[y(t)] =
Z
∞
e −sty(t)dt
0
Now as y(t) is a piecewise-defined defined function, it follows that e −sty(t) is also piecewise-defined:


06 t<1
 0
−st
e y(t) = e −stt
16t<2

 0
t>2
and
Z ∞
Z 1
Z 2
Z ∞
−st
−st
−st
e y(t)dt=
e y(t)dt +
e y(t)dt +
e −sty(t)dt
0
0
1
2
Z 1
Z 2
Z ∞
=
0dt +
e −sty(t)dt +
0 dt
0 Z
1
2
2
=0+
e −sttdt + 0
1
Z 2
=
e −sttdt
1
and using integration by parts with
du = e−st dt
e−st
dv = 1 dt
u=−
s
Z
e−st t 2 1 2 −st
+
e −s tt dt = −
e
dt
s 1
s
1
e−st 2
e−st t 2
+ −
= −
s2
s
1 1
1 1 −s
1 2 −2s
= 2+
e − 2+
e
s
s
s
s
1 1 −s
1 2 −2s
L[y(t)] = 2 +
e − 2+
e .
s
s
s
s
v=t
we have
Z
1
and therefore
2
2.1 Definition and existence of Laplace transforms
77
We now state a theorem that guarantees the existence of the Laplace transform of functions that are
of exponential order and that are piecewise continuous.
Theorem 2.15. (Existence) Let y(t) be a function such that |y(t)|<A ebt when t > t0 for some
constants A, b and t0 (that is, y(t) is of exponential order). Also let y(t) be piecewise continuous on
any finite interval c1 6 t 6 c2 where c1, c2 > 0. Then the Laplace transform of y(t) exists for s > b.
Proof. Not required.
Example 2.16. Show that the following functions are of exponential order and hence use Theorem
2.15 to prove that their Laplace transforms exist.2.2
i.
e5t
ii. sin 2t

0
iii. y(t) = t

0
06 t<1
16t<2
t>2
Answer:
i.
|e5t |<1.e5t for t > 0 and so the constants of Definition 2.9 are A = 1, b = 5 and t0 = 0. e5t is a
function that is continuous everywhere (see the graph on Figure 2.1) and is hence continuous
on every finite interval and is therefore piecewise continuous on every finite interval . So by
Theorem 2.15 the Laplace transform of e5t exists for s > 5, a fact we already know from
Example 2.5.
ii. |sin 2t| < 2 = 2e0t for t > 0 and so A = 2, b = 0 and t0 = 0. sin 2t is continuous everywhere and is
therefore piecewise continuous on every finite interval. So by Theorem 2.15 the Laplace transform of sin 2t exists for s > 0, which we already know from Example 2.6.
iii. From the graph of y(t) in Example 2.14 we see that |y(t)| < 2 for t > 0 and therefore |y(t)| <
2 = 2e0t for t > 0 and so A = 2, b = 0 and t0 = 0. If we consider any finite interval c1 6 t 6 c2
where c1, c2 > 0 then y(t) has at most two points of discontinuity in the interval c1 6 t 6 c2
depending on whether or not x = 1, 2 lie in the interval c1 6 t 6 c2. Hence y(t) is piecewise continuous on every finite interval. By Theorem 2.15 the Laplace transform of y(t) exists for s >
0.
Remark 2.17. It is possible to obtain the Laplace transform of functions that do not satify the
1 2.3
and the Dirac delta ‘function’ considered later. The
t
hypotheses of Theorem 2.15, for example √
main use of Theorem 2.15 is to guarantee the existence of the Laplace transform of standard functions such as ekt , cos kt, sinh kt, t5 + 4t3 − 3t + 1 and piecewise functions such as those in Examples
2.12 and 2.14.
2.2. Assume that we do not know the Laplace transforms of these functions have not been determined in Examples 2.5,
2.6 and 2.14. Note we do not actually find Laplace transforms by Theorem 2.15, we just show that the transform exists.
2.3. See Kreyszig [2] Section 6.1
78
Laplace transforms
2.2 Properties of Laplace transforms
2.2.1 Linearity property
The Laplace transform has the linearity property
Theorem 2.18. Let y1(t) and y2(t) be functions whose Laplace transforms exist and let c1 and c2 be
scalars. Then
L[c1 y1(t) + c2 y2(t)] = c1L[y1(t)] + c2L[y2(t)].
Proof. Using the definition of the Laplace transform
Z ∞
L[c1 y1(t) + c2 y2(t)] =
e−st c1 y1(t) + c2 y2 dt
0Z
Z ∞
∞
−st
= c1
e y1(t)dt + c2
e−sty1(t)dt
0
0
= c1L[y1(t)] + c2L[y2(t)]
Example 2.19. Use the linearity property of the Laplace transform to determine
i. L[t2 + 3t + 2]
ii. L[cosh kt] where k is a real constant.
Answer:
i.
L[t2 + 3t + 2] = L[t2] + 3L[t] + 2L[1]
1
1
2
= 3 +3 2 +2
s
s
s
3 2
2
= 3+ 2+
s
s
s
ii. By definition cosh kt =
e−kt + ekt
therefore
2
−kt
e + ekt
L[cosh kt] = L
2
1
1
= L[e−k t] + L[ek t]
2
2 1
1
1
1
=
+
2 s+k
2 s−k
1 (s − k) + (s + k)
=
(s + k)(s − k)
2
s
= 2
s − k2
2.2.2 The inverse Laplace transform
In the above sections, we saw that the Laplace transform changes a function y(t) in the t variable to
a function L[y(t)] = Y (s) in a new independent variable s. We now consider the reverse problem
Definition 2.20. Given some function Y (s), if there exists a function y(t) such that L[y(t)] =
Y (s) then we call y(t) the inverse Laplace transform of Y (s) and write
L−1[Y (s)] = y(t).
2
Example 2.21. Find the inverse Laplace transform of Y (s) = s2 + 4 .
79
2.2 Properties of Laplace transforms
Answer: This is done by recognizing the form of the function Y (s). We recognize that
L[sin kt] =
and therefore
L[sin 2t] =
k
s2 + k 2
2
s2 + 4
hence by the definition of the inverse Laplace transform
2
L−1 2
= sin 2t.
s +4
A useful fact is that the inverse Laplace transform has the linearity property
Lemma 2.22. Let Y1(s) and Y2(s) be functions whose inverse Laplace transforms exist. Then for
any scalars c1, c2
L−1[c1Y1(s) + c2Y2(s)] = c1L−1[Y1(s)] + c2L−1[Y2(s)].
Proof. Let the inverse Laplace transforms of Y1(s) and Y2(s) be y1(t) and y2(t) respectively, that is
L−1[Y1(s)] = y1(t),
L−1[Y2(s)] = y2(t).
(2.31)
Then by the definition of the Laplace inverse transform
L[y1(t)] = Y1(s),
L[y2(t)] = Y2(s)
(2.32)
and using the linearity property of the Laplace transform
L[c1 y1(t) + c2 y2(t)] = c1L[y1(t)] + c2L[y2(t)].
By substituting (2.32) we get
L[c1 y1(t) + c2 y2(t)] = c1Y1(s) + c2Y2(s)
and using the definition of the inverse transform
c1 y1(t) + c2 y2(t) = L−1[c1Y1(s) + c2Y2(s)].
Finally, using (2.31) we have
c1L−1[Y1(s)] + c2L−1[Y1(s)] = L−1[c1Y1(s) + c2Y2(s)].
s
Example 2.23. Find the inverse Laplace transform of Y (s) = s2 − 9
By the linearity property of the inverse Laplace transform
i
s
2
11
L−1[Y (s)] = L−1 2
+ 4+ 2
s −9 s
s +5
h
h s
2
= L−1 2
+ L−1 4 + L−1
s −9
s
2
11
+ s4 + s2 + 5 .
11
√
s2 + ( 5 )2
#
(2.33)
and we showed in previous examples that
y(t)
cosh kt
tn
sin kt
L[y(t)]
s
s2 − k 2
n!
sn+1
k
s2 + k 2
(2.34)
80
Laplace transforms
We use the linearity property again to rewrite the constants of (2.33)
#
h
h s
11
−1
−1
−1 2
−1
√
L [Y (s)] = L
+L
+L
s2 − 9
s4
s2 + ( 5 )2
√
h h
3!
5
s
2
11
= L−1 2 2 + L−1 4 + √ L−1 2 √
s −3
3!
s
so that the functions inside the square brackets of L−1[
table (2.34) and hence we have
5
s + ( 5 )2
#
] ‘match’ those in the right column of the
√
1
11
L−1[Y (s)] = cosh 3t + t3 + √ sin( 5 t).
3
5
The following well-known result on partial fractions is useful when finding inverse transforms:
Lemma 2.24.
p(s)
(Partial Fraction Expansion)Let
be a ratio of polynomials with real coefficients. If the
q(s)
denominator q(s) factors as
q(s) = k(s − a1)n1(s − a2)n2 (s − ak)nk(s2 + b1s + c1)m1 (s2 + b js + c j )m j
where the quadratic factors are irreducible then the following identity holds
A1,1
A1,2
A1,n1
p(s)
≡
+
+ +
2
q(s) s − a1 (s − a1)
(s − a1)n1
A2,1
A2,2
A2,n2
+
+
+ +
2
(s − a2)n2
s − a2 (s − a2)
Ak,1
Ak,2
A2,nk
+
+
+ +
2
(s − ak)nk
s − ak (s − ak)
B1,m s + C1,m1
B1,1s + C1,1
B1,2s + C1,2
+ 2
+
+ 2 1
(s + b1s + c1)m1
s + b1s + c1 (s2 + b1s + c1)2
(2.35)
Bj ,m s + C j ,mj
Bj ,1s + C j ,1
Bj ,2s + C j ,2
+ 2
+ 2
+ 2 j
2
(s + b js + c j )mj
s + b js + c j (s + b js + c j )
+ R(s)
where R(s) is a polynomial that is equal to zero if the degree of the polynomial p(s) is less than the
degree of the polynomial q(s) and Ar,s , Br,s and Cr,s are real constants for all subscripts r, s.
Proof. Not required.
Example 2.25. The partial fraction expansion of
2s2 + 5s + 1
takes the form
(s − 1)(s + 2)2(s − 5)3(s2 + 1)(s2 + 9)2
A
B
C
2s2 + 5s + 1
≡
+
+
(s − 1)(s + 2)2(s − 5)3(s2 + 1)(s2 + 9)2 s − 1 s + 2 (s + 2)2
D
E
F
+
+
+
s − 5 (s − 5)2 (s − 5)3
Gs + H Is + J Ks + L
+ 2
+ 2
+
s +1
s + 9 (s2 + 9)2
where A, B, C , D, E , F , G, H , I , J , K and L are constants to be determined.
81
2.2 Properties of Laplace transforms
Example 2.26. Find the inverse Laplace transform of Y (s) =
s2 − s + 9
.
s3 + 9s
s2 − s + 9
As it is, we do not immediately recognize the expression s3 + 9s as the Laplace transform of a funcs2 − s + 9
tion that we know. So we use partial fractions to write s3 + 9s
do recognize. We first factorize the denominator
as a sum of expressons in s that we
s2 − s + 9 s2 − s + 9
=
s3 + 9s
s(s2 + 9)
then by the formula (2.35) the partial fraction expansion in this case is
s2 − s + 9 A Bs + C
≡ + 2
s(s2 + 9)
s
s +9
where A, B and C need to be determined. We do this by first clearing denominators
s2 − s + 9 ≡ A(s2 + 9) + s(Bs + C).
(2.36)
We need to find three constants (namely A, B and C) so shall need three equations with variables
A, B and C which may be obtained by either
•
substituting values – for example putting s = 0 into (2.36) gives
9 = A(0 + 9) + 0(B.0 + C)
⇒
A=1
•
comparing coefficients of like powers – rearranging the right side of (2.36) gives
s2 − s + 9 ≡ (A + B)s2 + Cs + 9A
and comparing the coefficients of s2 in (2.37) gives
A + B = 1,
while comparing the coefficients of s gives
C =−1
and therefore we have three equations in the variables A, B and C
A=1
A+B =1
C =−1
and solving these give A = 1, B = 0 and C = − 1. Hence
2
s −s+9
L−1[Y (s)] = L−1
s(s2 + 9)
1
1
= L−1 − 2
s s +9
1
1
−1
−1
−L
=L
s2 + 9
s
1
3
1
= L−1
− L−1 2
s
s +9
3
1
= 1 − sin(3t)
3
which is our answer.
(2.37)
82
Laplace transforms
2.2.3 Shifting the s variable; shifting the t variable
Given a function f (x), recall that the graph of the function f (x − a) is ‘shifted’ horizontally by a
units. If a > 0 then this horizontal shift is to the right; else if a < 0 the shift is to the left. The following graph is an example of a shift to the right when f (x) = sin(x) and a = 2.
1
0.5
sin(x)
0
-2
-1
0
1
2
3
4
5
6
-0.5
-1
sin(x-2)
We consider shifting in the context of Laplace transforms. Recall that the Laplace transform
changes functions y(t) in the t variable to corresponding functions Y (s) in the variable s
so we can shift the s variable or shift with respect to the t variable.
Shifting the s-variable
Let Y (s) be a function of s with inverse Laplace transform
L−1[Y (s)] = y(t).
If we shift Y (s) then we obtain a new function Y (s − a) in the variable s. The inverse Laplace transform of this new function Y (s − a) is related to the inverse transform of Y (s) by the formula
L−1[Y (s − a)] = eatL−1[Y (s)].
(2.38)
We state and prove this result in the following
Lemma 2.27. Let Y (s) have inverse Laplace transform L−1[Y (s)], and let Y (s − a) be the function
obtained by shifting Y (s). Then
L−1[Y (s − a)] = eatL−1[Y (s)].
Proof. Let L−1[Y (s)] = y(t). Then
L[eaty(t)] =
=
Z
∞
Z 0∞
e−st eaty(t)) dt
(2.39)
e−(s−a)ty(t)dt
0
Y (s) = L[y(t)] and by the definition of the Laplace transform
Z ∞
e−sty(t)dt.
Y (s) =
(2.40)
0
Replacing s by s − a on both sides of (2.40) gives
Z ∞
Y (s − a) =
e−(s−a)ty(t)dt.
(2.41)
0
But this is identical to the second integral of (2.39) and therefore
L[eaty(t)] = Y (s − a)
(2.42)
83
2.2 Properties of Laplace transforms
which implies
L−1[Y (s − a)] = eaty(t)
⇒ L−1[Y (s − a)] = eatL−1[Y (s)].
Equation (2.42) above will also be useful so we state it in the following
Corollary 2.28. If the function y(t) has Laplace transform Y (s) then
L[eaty(t)] = Y (s − a).
Lemma 2.27 and Corollary 2.28 are particularly useful in finding inverse Laplace transforms.
Example 2.29.
1
−1
i. Find L
s2 − 2s + 10
h
1
ii. Find L−1 (s + a)n+1 when n = 1, 2, 3
Answer:
i. First complete the square of the quadratic2.4
s2 − 2s + 10 ≡ (s − 1)2 + 9.
We need to determine
L
−1
1
1
−1
=L
.
s2 − 2s + 10
(s − 1)2 + 9
1
3
= L−1
3
(s − 1)2 + 9
Notice that this is similar to
L[sin 3t] =
shifting this by using (2.42) we have
3
,
s2 + 9
3
(s − 1)2 + 9
3
1
1
= L−1
L−1 2
(s − 1)2 + 9
3
s − 2s + 10
1 t
= e sin 3t.
3
L[etsin 3t] =
and therefore
ii. Recall from Example 2.7
L[tn] =
n!
when n = 1, 2, 3
sn+1
From formula (2.42) with a replaced by − a
L[e−attn] =
n!
(s + a)n+1
and from the linearity property of Laplace transforms
L[
therefore
1 −at n
1
1
e t ] = L[e−attn] =
n!
n!
(s + a)n+1
1
1
−1
L
= e−attn.
(s + a)n+1
n!
b
b2
2.4. The formula for completing the square is ax2 + bx + c ≡ a(x + 2a )2 + c − 4a
84
Laplace transforms
Shifting the t-variable
When shifting the t variable, a function that is useful is the Heaviside step function.
Definition 2.30.
The Heaviside step function H(t − t0) is defined as
0
when t < t0
.
H(t − t0) =
1
when t > t0
1
1
t
0
t0
Figure 2.2. Graph of H(t − t0)
Notice that H(t − t0) is a piecewise defined function and using the definition of the Laplace transform as in Example 2.14 we can determine L[H(t − t0)].
Lemma 2.31. The Laplace transform of the Heaviside step function is L[H(t − t0)] =
e−st0
.
s
Proof.
L[H(t − t0)] =
=
∞
Z
Z0
t0
0
e−stH(t − t0)dt
Z ∞
−st
e−s t(1)dt
e (0)dt +
t0
from the definition of H(t − t0). Therefore
Z
∞
e−stdt
Z R
e −st dt
= lim
R→∞ t0
−st R
e
= lim −
s
R→∞
t0
−st0
e
e−sR
= lim
−
s
s
R→∞
e−s R
e−st0
− lim
=
s
s
R→∞
−s t0
e
=
s
L[H(t − t0)] =
as limR→∞ e−sR = 0.
t0
85
2.2 Properties of Laplace transforms
Example 2.32. Sketch the graphs of the following functions
i. y1(t) = sin t when t > 0
ii. y2(t) = sin (t − 2) when t > 2
iii. y3(t) = H(t − 2)sin(t − 2) when t > 0
Answer: The graphs of i) and ii) are
1
1
0.5
0.5
0
0
0
-0.5
1
2
3
4
5
0
1
2
3
4
5
6
7
8
9
-0.5
y1(t)
y2(t)
-1
-1
The graph of y2(t) is obtained from y1(t) by shifting 2 units to the right. Notice that y2(t) is not
defined for 0 6 t < 2. y3(t) is the product of functions and we use the following table to determine its
graph
t interval H(t − 2) H(t − 2)sin(t − 2)
06t<2
0
0
t>2
1
sin(t − 2)
so we see that
H(t − 2)sin(t − 2) =
0
sin(t − 2)
when 0 6 t < 2
when t > 2
and therefore the graph of y3(t) is obtained y1(t) by shifting 2 units to the right and taking the
graph to be zero on the interval 0 6 t < 2. Notice that, unlike y2(t), y3(t) is defined for 0 6 t < 2 and
the answer for part iii) is
1
0.5
0
0
1
2
3
4
5
6
7
8
9
-0.5
y3(t)
-1
We saw in Example 2.32, that H(t − 2)sin(t − 2) is a ‘t-shift’ of sin(t − 2). Below we give a result
that expresses the Laplace transform of a shifted function H(t − t0)y(t − t0) in terms of the Laplace
transform of the original function
Theorem 2.33. Let L[y(t)] = Y (s). Then
L[H(t − t0)y(t − t0)] = e−st0Y (s).
Proof.
L[H(t − t0)y(t − t0)] =
=
∞
Z
Z0
0
t0
e−stH(t − t0)y(t − t0)dt
Z ∞
e−st(1)y(t − t0)dt
e−st(0)y(t − t0)dt +
t0
86
Laplace transforms
from the definition of H(t − t0) in Definition 2.30. Therefore
Z ∞
e−sty(t − t0)dt.
L[H(t − t0)y(t − t0)] =
t0
Now use the change of variables T = t − t0 in the integral on the right to get
Z ∞
L[H(t − t0)y(t − t0)] =
e−s(T +t0) y(T )dT
0
Z ∞
= e−st0
e−sTy(T )dT
0
But T is a ‘dummy variable’ in the above integral that is, the integral takes the same value regardless
of the name of the independent variable:
Z ∞
Z ∞
e−sTy(T )dT =
e−sty(t)dt.
0
0
Therefore
Z
L[H(t − t0)y(t − t0)] = e−st0
∞
e−sty(t)dt
0
= e−s t0L[y(t)]
= e−s t0Y (s)
which is the desired result.
Using the definition of the inverse Laplace transform and Theorem 2.33 we clearly have
Corollary 2.34. Let L[y(t)] = Y (s). Then
L−1[e−st0Y (s)] = H(t − t0)y(t − t0).
Example 2.35. Sketch the graph of L
h
−1
e−3s
s2 + 1
i
.
1
e−3s
is of the form e−st0Y (s) with e−st0 = e−3s and Y (s) = 2
. Now
2
s +1
s +1
1
⇒ y(t) = L−1[Y (s)] = sin(t)
Y (s) = 2
s +1
and by Corollary 2.34
−3s −1 e
L
= H(t − 3)y(t − 3)
s2 + 1
= H(t − 3)sin(t − 3)
Answer: Notice that
Using the definition of H(t − t0) in Definition 2.30
0
H(t − 3)sin(t − 3) =
sin(t − 3)
when 0 6 t < 3
when t > 3
and following Example 2.32 the graph of H(t − 3)sin(t − 3) is
1
0.5
t
0
0
-0.5
-1
1
2
3
4
5
6
7
8
9
87
2.2 Properties of Laplace transforms
2.2.4 Laplace transform of derivatives
The Laplace transform of the derivative y ′(t) can be expressed in terms of the Laplace transform of
the original function y(t). This is essentially done by using integration by parts
Z ∞
L[y ′(t)] =
e−sty ′(t)dt
0
Z R
e−sty ′(t)dt
= lim
R→∞
now let
0
v = e−st
dv = − se−st dt
and so
L[y ′(t)] = lim
R→∞
= lim
R→∞
Z
R
du = y ′(t) dt
u = y(t)
e−s ty ′(t)dt
0
e
−st
−sR
y(t)
R
0
+s
Z
R
e
−st
y(t)dt
0
0
!
Z
R
= lim e
y(R) − e y(0) + s lim
e
R→∞
R→∞ 0
= lim e−sRy(R) − y(0) + sL[y(t)].
(2.43)
−st
y(t)dt
R→∞
and provided limR→∞ e−sRy(R)=0 (as will be the case if y(t) is of exponential order) then we have
that
L[y ′(t)] = sL[y(t)] − y(0)
(2.44)
which expresses the Laplace transform of y ′(t) in terms of the Laplace transform of y(t). Note that
the argument (2.43) is not a proof, we have ignored some continuity issues. The full proof, given in
Ramkissoon [1], is not required for this course. We give a statement of the hypotheses required for
(2.44) in the following theorem.
Theorem 2.36. If y(t) is continuous and of exponential order and y ′(t) is piecewise continuous on
every finite interval then
L[y ′(t)] = sL[y(t)] − y(0).
Proof. Not required.
By two applications of Theorem 2.36 we can determine the Laplace transform of y ′′(t) in terms of
the transform of y(t):
h
′
L[y ′′(t)] = L y ′(t))
= sL[y ′(t)] − y ′(0)
= s (sL[y(t)] − y(0)) − y ′(0)
using (2.44) with y replaced by y ′
using (2.44) as it is
which gives us
L[y ′′(t)] = s2L[y(t)] − sy(0) − y ′(0)
(2.45)
Similarly, one can apply Theorem 2.36 n times to obtain the Laplace transform of the nth derivative
y (n)(t) in terms of the transform of y(t):
L[y(n)(t)] = snL[y(t)] − sn−1 y(0) − sn−2 y ′(0) − sn−3 y ′′(0) − sy (n−2)(0) − y (n−1)(0)
(2.46)
The formulae (2.45) and (2.46) require certain continuity conditions to hold. We give a statement of
these conditions in the following theorem.
88
Laplace transforms
Theorem 2.37. If y(t), y ′(t), y ′′(t), , y (n−1)(t) are continuous and of exponential order and y(n)(t)
is piecewise continuous on every finite interval then
L[y (n)(t)] = snL[y(t)] − sn−1 y(0) − sn−2 y ′(0) − sn−3 y ′′(0) − sy (n−2)(0) − y (n−1)(0).
Proof. Not required.
Example 2.38. Use the formula for the Laplace transform of a derivative to determine
i. L[cos(kt)]
ii. L[sinh(kt)]
Answer:
i. Let y(t) = sin(kt). Then y ′(t) = k cos(kt) and applying Theorem 2.36 we have
L[y ′(t)] = sL[y(t)] − y(0)
⇒ L[k cos(kt)] = sL[sin(kt)] − sin(0)
⇒ kL[ cos(kt)] = sL[sin(kt)] − sin(0)
where we used the linearity property of Theorem 2.18 to move the k outside of the Laplace
transform. Then
k
kL[ cos(kt)] = s 2
s + k2
k
as we know L[sin kt] = s2 + k 2 from Example 2.6. Therefore
L[ cos(kt)] =
s
s2 + k 2
ii. Recall that the hyperbolic cosine cosh(t) and hyperbolic sine sinh(t) are defined as
e kt + e−kt
2
e kt − e−kt
sinh kt =
2
cosh kt =
and it is easy to check from these definitions that
(cosh kt) ′ = k sinh kt , (sinh kt) ′ = k cosh kt
cosh(0) = 1 , sinh(0) = 0
Let y(t) = cosh kt. Then y ′(t) = k sinh(kt) and applying Theorem 2.36 we have
L[k sinh(kt)] = sL[cosh kt] − cosh(0)
s
and as we know L[cosh kt] = s2 − k 2 from Example 2.19
kL[sinh(kt)] = s
and so
s
−1
s2 − k 2
s2 − (s2 − k 2)
⇒ kL[sinh(kt)] =
s2 − k 2
L[sinh(kt)] =
k
.
s2 − k 2
(2.47)
89
2.3 Applications and more properties of Laplace transforms
2.3 Applications and more properties of Laplace transforms
2.3.1 Solving differential equations using Laplace transforms
We can use the results of the previous sections to solve differential equations, in particular initial
value problems, that is, differential equations with specified initial values. We illustrate the application of Laplace transforms to initial value problems in the following examples.
Example 2.39. Solve the differential equation
y ′′ − 3y ′ + 2y = e3t
(2.48)
subject to the initial conditions y(0) = 0, y ′(0) = 0.
Answer: Take the Laplace transform of both sides of the given differential equation
L[y ′′ − 3y ′ + 2y] = L[e3t]
1
⇒ L[y ′′] − 3L[y ′] + 2L[y] =
s−3
Using formulae (2.44), (2.45) for the transform of a derivative we have
s2L[y(t)] − sy(0) − y ′(0) − 3(sL[y(t)] − y(0)) + 2L[y(t)] =
Substituting the given initial conditions y(0) = 0, y ′(0) = 0 gives
s2L[y(t)] − 3sL[y(t)] + 2L[y(t)] =
For convenience let L[y(t)] = Y (s)
s2Y (s) − 3sY (s) + 2Y (s) =
Now solve for Y (s)
(s2 − 3s + 2)Y (s) =
Y (s) =
1
s−3
1
s−3
1
s−3
1
s−3
1
(s2 − 3s + 2)(s − 3)
(2.49)
Our required answer is y(t) = L−1[Y (s)]. We use the partial fraction expansion method (as in
Example 2.26) to find the inverse transform of (2.49). We first factorize the denominator of Y (s)
1
1
=
(s2 − 3s + 2)(s − 3) (s − 1)(s − 2)(s − 3)
then by the formula (2.35) the partial fraction expansion is
A
B
C
1
≡
+
+
(s − 1)(s − 2)(s − 3) s − 1 s − 2 s − 3
where A, B and C need to be determined. Clear denominators
1 ≡ A(s − 2)(s − 3) + B(s − 1)(s − 3) + C(s − 1)(s − 2)
(2.50)
In this case it is easiest to substitute values into (2.50) in order to determine A, B and C. Substituting values s = 1, s = 2 and s = 3 into (2.50) gives
90
Laplace transforms
s=1
1 = A( − 1)( − 2) + B(0)( − 2) + C(0)( − 1)
1
⇒ A=
2
s=2
1 = A(0)( − 1) + B(1)( − 1) + C(1)(0)
⇒ B =−1
s=3
1 = A(1)(0) + B(2)(0) + C(2)(1)
1
⇒ C=
2
Therefore
1
y(t) = L [Y (s)] = L
(s2 − 3s + 2)(s − 3)
1
= L−1
(s − 1)(s − 2)(s − 3)
" 1
#
1
(
−
1)
2
= L−1
+
+ 2
s−1 s−2 s−3
1
1
1
1
1
− L−1
+ L−1
= L−1
s−2
2
s−1
s−3
2
1 t
1 3t
2t
= e −e + e
2
2
−1
−1
and so the solution to this initial value problem is
1
1
y(t) = et − e2t + e3t
2
2
It is possible to solve the initial value problem of Example 2.39 by other methods of solving differential equations such as D-operators and power series. One value of using Laplace transforms has over
these other methods is that the inhomogeneous term, usually on the right side of the differential
equation (for example the e3t in equation (2.48)) can be a piecewise defined function as we see in the
following example.
Example 2.40. Solve the differential equation
x ′′(t) + x(t) = F (t)
subject to the initial conditions x(0) = 0, x ′(0) = 1 and where F (t) is the piecewise defined function

π

1
when 0 6 t <
2
F (t) =
π

0
when t >
2
Answer: Take the Laplace transform of both sides of the given differential equation
L[x ′′ + x] = L[F (t)]
and so we need to determine L[F (t)]. The graph of F (t) is
(2.51)
91
2.3 Applications and more properties of Laplace transforms
11
t
0
pi/2
One way to determine the Laplace transform of a piecewise defined function is to do so from first
principles, that is, using the definition of the Laplace transform as in Example 2.14. Another way is
to use Heaviside functions. Consider the following table
t interval
π
2
π
t>
2
06t<
and therefore
π
π
H(t − ) 1 − H(t − )
2
2
F (t)
0
1
1
1
0
0
π
F (t) = 1 − H(t − )
2
h
π i
⇒ L[F (t)] = L 1 − H(t − )
2
h
π i
= L[1] − L H(t − )
2
sπ
−2
1 e
= −
s
s
using Lemma 2.31. Substituting L[F (t)] into (2.51) we have
sπ
−
1 e 2
L[x + x] = −
s
s
sπ
−2
1 e
′′
L[x ] + L[x] = −
s
s
′′
⇒
Using formula (2.45) for the transform of the second derivative we have
sπ
−
1 e 2
s L[x(t)] − sx(0) − x (0) + L[x(t)] = −
s
s
2
′
Substituting the given initial conditions x(0) = 0, x ′(0) = 1 gives
sπ
−
1 e 2
s L[x(t)] − 1 + L[x(t)] = −
s
s
For convenience let L[x(t)] = X(s)
sπ
−2
1
−
e
s2X(s) − 1 + X(s) =
s
and solve for X(s)
sπ
−
1−e 2
2
(s + 1)X(s) =
+1
s
sπ
−
1
1−e 2
+
X(s) =
s(s2 + 1) s2 + 1
2
(2.52)
At this point we use partial fractions to expand
sπ
−
1−e 2
s(s2 + 1)
(2.53)
92
Laplace transforms
In this case it may be best to write (2.53) as
sπ
−2
1
1−e
s(s2 + 1)
and apply partial fractions to
(2.54)
1
:
s(s2 + 1)
1
A Bs + C
+ 2
s
s +1
1 ≡ A(s2 + 1) + s(Bs + C)
1 ≡ (A + B)s2 + Cs + A
s(s2 + 1)
⇒
⇒
≡
and by comparing coefficients of 1, s and s2 we obtain, respectively, three equations
and so
and therefore
A=1
C =0
A+B =0
1
1
s
≡ −
.
s(s2 + 1) s s2 + 1
sπ
sπ
−2
−2
s
1
1
−
1−e
=
1−e
s s2 + 1
s(s2 + 1)
sπ
sπ
−
−
1
s
e 2
se 2
= − 2
−
+ 2
s s +1
s
s +1
and this is the required partial fraction expansion of (2.54). Subtituting this into X(s) in (2.52) we
have
sπ
sπ
−
−
1
1
s
e 2
se 2
X(s) = + 2
−
−
+ 2
s s + 1 s2 + 1
s
s +1
and therefore


sπ
sπ
−2
−2
1
s
e
se
1

x(t) = L−1[X(s)] = L−1 + 2
−
−
+ 2
s s + 1 s2 + 1
s
s +1




sπ
sπ
−2
−2
1
s
1
se
e


+ L−1 2
− L−1 2
− L−1
+ L−1 2
= L−1
s +1
s +1
s
s +1
s
π
π
π
= 1 + sin t − cos t − H(t − ) + H(t − )cos(t − )
2
2
2
and so the solution to this initial value problem is
π
π
π
x(t) = 1 + sin t − cos t − H(t − ) + H(t − )cos(t − ).
2
2
2
(2.55)
So when the inhomogeneous part F (t) of our differential equation is piecewise defined, we obtain
a solution to the differential equation that is also piecewise defined as we can see by using the definition of the Heaviside function to write (2.55) as

π

 1 + sin t − cos t
when 0 6 t <
2
x(t) =
π
π

 1 + sin t − cos t − 1 + cos(t − ) when t >
2
2
It is interesting to note that while the inhomogeneous part F (t) of the differential equation of
π
π
Example 2.40 is not continuous at t = 2 (there is a break in the graph of F (t) at t = 2 ), the solution
π
x(t) given in (2.55) is actually continuous and in fact differentiable at t = 2 as we can see from the
graph of x(t)
93
2.3 Applications and more properties of Laplace transforms
x(t)
2
1
t
0
pi/2
-1
-2
π
π
where we have drawn the interval 0 6 t < in solid and the interval t > in dashes. Notice both
2
2
π
pieces meet smoothly at t = . Also notice from the graph that x(t) satisfies the initial conditions
2
x(0) = 0, x ′(0) = 1.
2.3.2 Solving simultaneous linear differential equations using
the Laplace transform
We briefly discuss an example that motivates how simultaneous differential equations arise. This
example serves only to help illustrate simultaneous differential equations and its details are not
required for this course.
Example 2.41. Consider a farm with mice and cats. The mouse population changes with time t
and denote this population by M (t). Similarly denote the cat population by C(t). Now the rate
M ′(t) at which the mouse population changes at time t depends on two things
−
the number of mice alive at time t (that is M (t)), as more mice present means a greater
mouse reproduction rate.
−
the number of encounters between cats and mice as the population of mice decreases at these
encounters. For simplicity we assume the number of encounters is proportional to the product
M (t)C(t).
Hence M ′(t) is proportional to M (t) and M (t)C(t) and we have the differential equation
M ′(t) = aM (t) − bM (t)C(t)
where a and b are constants such that a, b > 0. Notice that this differs from the differential equations
we have considered so far as there are two dependent variables M , C and one independent variable t.
Similarly the rate C ′(t) at which the cat population changes depends on
−
the number of encounters between cats and mice as encounters means food for the cats and
therefore an increased survival rate C ′(t) of cats
−
the more cats present means less food for each individual cat and therefore an increased
death rate C ′(t) of cats
and we have the differential equation
where d, e > 0. The two equations
C ′(t) = − dC(t) + eM (t)C(t)
M ′(t) = aM (t) − bM (t)C(t)
C ′(t) = − dC(t) + eM (t)C(t)
form an example of simultaneous differential equations in which the dependent variables interact2.5.
2.5. This particular system of differential equations is called a Lotka-Volterra system and is used in predator-prey models.
94
Laplace transforms
We now illustrate the solution of simultaneous linear differential equations by using the Laplace
transform.
Example 2.42. Use the Laplace transform to solve the following simultaneous linear differential
equations
y1′ = − y1 + y2
y2′ = − y1 − y2
subject to the initial conditions y1(0) = 1 and y2(0) = 0.
Answer: Take the Laplace transform of both equations
L[y1′ ] = − L[y1] + L[y2]
L[y2′ ] = − L[y1] − L[y2]
Using formulae (2.44) for the transform of a derivative we have
sL[y1] − y1(0) = − L[y1] + L[y2]
sL[y2] − y2(0) = − L[y1] − L[y2]
Substituting the initial conditions y1(0) = 1 and y2(0) = 0 gives
sL[y1] − 1 = − L[y1] + L[y2]
sL[y2] = − L[y1] − L[y2]
For convenience let L[y1(t)] = Y1(s) and L[y2(t)] = Y2(s)
sY1(s) − 1 = − Y1(s) + Y2(s)
sY2(s) = − Y1(s) − Y2(s)
and we can rewrite these two equations as
− 1 = − (s + 1)Y1(s) + Y2(s)
(2.56)
0 = − Y1(s) − (s + 1)Y2(s)
(2.57)
We eliminate Y2(s). Multiplying equation (2.56) by s + 1 we have
− (s + 1) = − (s + 1)2Y1(s) + (s + 1)Y2(s)
(2.58)
and adding equations (2.57) and (2.58) gives
From equation (2.57) we know
− (s + 1) = − ( (s + 1)2 + 1) Y1(s)
s+1
⇒ Y1(s) =
(s + 1)2 + 1
Y1(s)
s+1
1
⇒ Y2(s) = −
(s + 1)2 + 1
Y2(s) = −
Recall that
L[cos t] =
s
s2 + 1
L[sin t] =
1
s2 + 1
Shifting these along the s-axis gives
L[e−tcos t] =
and so
s+1
(s + 1)2 + 1
y1(t) = L−1[Y1(s)]
s+1
= L−1
(s + 1)2 + 1
−t
= e cos t
L[e−tsin t] =
1
(s + 1)2 + 1
y2(t) = L−1[Y2(s)]
1
= L−1 −
(s + 1)2 + 1
−t
= − e sin t
95
2.3 Applications and more properties of Laplace transforms
and the solution to our simultaneous linear differential equations is
y1(t) = e−tcos t
y2(t) = − e−tsin t.
2.3.3 Convolution and Integral equations
Recall from Lemma 2.22 that the inverse Laplace transform of a sum is the sum of inverse transforms
L−1[Y1(s) + Y2(s)] = L−1[Y1(s)] + L−1[Y2(s)].
However, a similar statement does NOT hold for products. In general
L−1[Y1(s)Y2(s)]
L−1[Y1(s)]L−1[Y2(s)]
(2.59)
1
1
and Y2(s) = . Then
s
s
1
1
L−1[Y1(s)]L−1[Y2(s)] = L−1
L−1
s
s
as the following example demonstrates: consider Y1(s) =
1 1
L−1[Y1(s)Y2(s)] = L−1
s s
−1 1
=L
s2
=t
= 1. 1
=1
and clearly this is a case of (2.59).
In order to obtain a formula for L−1[Y1(s)Y2(s)] where Y1(s) and Y2(s) are Laplace transforms, we
shall need the following
Definition 2.43. The convolution of f (t) and g(t) is denoted as f ∗ g and is defined as
Z t
(f ∗ g)(t) =
f (t − β)g(β)dβ.
0
Example 2.44. Find the convolution f ∗ g when f (t) = t2 and g(t) = t, that is, find t2 ∗ t.
Answer: Replacing the variable t with t − β in f (t) = t2 gives
and similarly we obtain g(β) = β. Then
and so
f (t − β) = (t − β)2
(f ∗ g)(t) = t2 ∗ t
Z t
=
(t − β)2 β dβ
0
Z t
=
(t2 − 2tβ + β 2)β dβ
0
Z t
=
(t2 β − 2tβ 2 + β 3) dβ
0
2 2
2tβ 3 β 4 t
tβ
−
+
=
3
4 0
2
t4 2t4 t4
+
= −
3
4
2
4
t
=
12
t2 ∗ t =
t4
.
12
Notice from Definition 2.43 and Example 2.44 that if f , g are functions of t then f ∗ g is again a
function of t. We now give a formula for L−1[Y1(s)Y2(s)] in the theorem below.
96
Laplace transforms
Theorem 2.45. (Convolution Theorem) If Y1(s) = L[y1(t)] and Y2(s) = L[y2(t)] then
L−1[Y1(s)Y2(s)] = y1 ∗ y2
or equivalently
L[y1 ∗ y2] = L[y1(t)]L[y2(t)].
Proof. Not required.
Example 2.46. Find L
Answer: We can split
h i
−1
2
s5
by using the Convolution Theorem.
2
as a product in several ways. Let us choose one such way
s5
Y1(s) =
and using the formula L[tn] =
n!
sn+1
2
s3
Y2(s) =
y1(t) = L−1[Y1(s)] = t2
1
.
s2
y2(t) = L−1[Y2(s)] = t
Then using the Convolution Theorem
2 1
2
L−1 5 = L−1 3 2
s s
s
= L−1[Y1(s)Y2(s)]
= y1 ∗ y2
= t2 ∗ t
and from Example 2.44
t2 ∗ t =
therefore
t4
12
2
t4
L−1 5 = .
s
12
h i
2
, we can use the
Of course, we did not need to use the Convolution Theorem to determine L
s5
n!
formula L[tn] = n+1 directly. On the other hand, examples in which the Convolution Theorem is
s
needed are given by integral equations.
−1
Definition 2.47. An equation which contains the dependent unknown variable under an integral
operator is called an integral equation.
We will be interested in those integral equations that can be solved by using Laplace tranforms and
the Convolution theorem.
Example 2.48. Use the Laplace transform to solve the integral equation
Z t
y(t) = 4t − 3
sin(t − β)y(β)dβ.
0
Answer: Take the Laplace transform of both sides of the integral equation
Z t
sin(t − β)y(β)dβ .
L[y(t)] = L[4t] − 3L
0
Notice that the integral
Z
0
t
sin(t − β)y(β)dβ
(2.60)
2.3 Applications and more properties of Laplace transforms
97
is in the form of a convolution, that is
Z t
sin(t − β)y(β)dβ = sin t ∗ y(t)
0
and from the Convolution Theorem
Z t
sin(t − β)y(β)dβ = L[sin t ∗ y(t)] = L[sin t]L[y(t)].
L
(2.61)
0
Substituting (2.61) into (2.60) we have
and denoting L[y(t)] = Y (s)
and using partial fractions
L[y(t)] = L[4t] − 3L[sin t]L[y(t)]
4
1
Y (s) = 2 − 3 2
Y (s)
s
s +1
3
4
⇒ Y (s) 1 − 2
= 2
s +1
s
2
s +4
4
⇒
Y (s) 2
= 2
s +1
s
4(s2 + 1)
⇒
Y (s) = 2 2
s (s + 4)
4(s2 + 1) A B Cs + D
≡ + + 2
s2(s2 + 4) s s2
s +4
⇒ 4s2 + 4 ≡ As(s2 + 4) + B(s2 + 4) + (Cs + D)s2
⇒ 4s2 + 4 ≡ (A + C)s3 + (B + D)s2 + 4As + 4B
and by comparing coefficients of 1, s , s2 and s3 we obtain, respectively, four equations
4B = 4
4A = 0
B+D=4
A+C =0
Solving these gives A = 0, B = 1, C = 0, D = 3 and therefore
Y (s) =
3
1
+
s2 s2 + 4
and so the answer to our integral equation is y(t) where
3
−1
−1
−1 1
+L
y(t) = L [Y (s)] = L
s2 + 4
s2
1
2
3
= L−1 2 + L−1 2
s
s + 22
2
3
= t + sin(2t).
2
2.3.4 Dirac’s delta function
Dirac’s delta function δ(t) is not actually a function of the variable t. It is a distribution. The
theory of distributions belongs to a branch of mathematics called functional analysis. We shall discuss the proof of the formula
L[δ(t)] = 1
(2.62)
which can more generally be stated as
L[δ(t − t0)] = e −st0
where t0 > 0
(2.63)
98
Laplace transforms
however this proof is not required for this course. Knowledge and application2.6of the two formulae
(2.62) and (2.63) is the only requirement of this subsection.
In functional analysis, a functional is a map from a set of functions to a set of numbers.
Example 2.49. Let C denote the set of continuous functions and R denote the set of real numbers.
Then
i. the map φ: C → R defined by
φ(f ) =
Z
b
f (t)dt
a
is a functional as integrating any continuous function f (t) over a finite interval [a, b] gives a
number.
ii. the map σ: C → R defined by
σ(f ) = f (0)
is also a functional. Notice that σ associates to any continuous function f (x) the value of
that function at t = 0.
One can define functionals on many sets of functions, in the Example 2.49 above we described two
different functionals on a particular set of functions – the set of continuous functions C. To define
the Dirac delta function we shall need another specific set of functions – the set of test functions.
Denote the set of test functions as D.
Definition 2.50. A test function on R is a function that is identically zero outside a sufficiently
large interval − c < t < c and that has derivatives of any order.
Recall that as any differentiable function is necessarily continuous, it follows that D ⊂ C, that is the
set of test functions on R is contained in the set of continuous functions on R.
Example 2.51. The function

 − 12
f (t) = e 1−t
0
when − 1 < t < 1
when t 6 − 1 or t > 1
is an example of a test function. From the graph of f (t)
f(t)
0
-1
0
1
it is clear that f (t) is identically zero outside − 1 < t < 1. Notice that f (t) is a continuous function.
Definition 2.52. Let D denote the set of all test functions on R. Then a functional that has
domain equal to D is called a distribution on R.
We can now give a precise definiton of the Dirac delta function δ(t).
2.6. see Example 2.63
99
2.3 Applications and more properties of Laplace transforms
Definition 2.53.
The Dirac delta function δ(t) is the distribution δ(t): D → R that is defined as
δ(t)(φ) = φ(0).
So the Dirac delta function δ(t) maps any test function φ in the set D to a corresponding number
φ(0). In other words, δ(t) takes any test function φ and evaluates that function at t = 0. More generally, we can define the ‘shifted’ Dirac delta function δ(t − t0).
Definition 2.54.
The Dirac delta function δ(t − t0) is the distribution δ(t − t0): D → R that is defined as
δ(t − t0)(φ) = φ(t0)
where t0 > 0 is a constant.
Notice by letting t0 = 0, Definition 2.54 reduces to Definition 2.53.
We now give a more intuitive but less technically accurate description of the Dirac delta function
δ(t − t0). Consider the following function

1

n when t0 6 t 6 t0 + n
fn(t) =
(2.64)
0
otherwise
where n = 1, 2, 3, is a parameter that can vary. The graph of fn(t) is
fn(t)
n
t
t0
t0+1/n
and we see that the graph of fn(t) takes the form of a pulse. Notice that the area under this graph is
1
n(t0 + n − t0) = 1. Now as the parameter n gets larger, the ‘pulse’ fn(t) gets narrower and higher.
We see this in the graphs below illustrating the cases n = 1, 2, 3 respectively:
f1(t)
f2(t)
f3(t)
3
2
1
t
t0+1
t
t0+1/2
t
t0+1/3
Notice that the area under the graph of each f1, f2 and f3 is 1.
Intuitively, as n → ∞ the functions fn tend to the Dirac delta function, that if it were to exist as
a function, would take the form of a pulse with zero width, infinite height and an area of 1.
100
Laplace transforms
The above intuitive description of the Dirac delta function is not precise as the pointwise limit of
the functions fn(t) as n → ∞
lim fn(t)
n→∞
does not exist. If the limit limn→∞ fn(t) were to exist then for any value t = a, the sequence of numbers fn(a) would converge. However, when t = t0 the sequence of numbers f1(t0), f2(t0), f3(t0), is
equal to 1, 2, 3, which is a sequence that does not converge. Therefore the pointwise limit
limn→∞ fn(t) does not exist.
However, it turns out that the theory of distributions was developed to handle such situations as
taking the limit of the sequence fn above. In the theory of distributions, each function fn can be
converted into a corresponding distribution which is denoted by Tfn . Then one takes the limit of
these distributions – this is actually a limit of integrals. In this particular case, the limit of the Tfn
as n → ∞ does exist and is in fact equal to the distribution δ(t − t0) as defined in Definition 2.54. It
is in this sense of distributions that the above sequence fn converge to the Dirac delta δ(t − t0).
The following definition describes how a function fn is converted to a distribution Tfn.
R b
Definition 2.55. Let f (t) be a function that is locally integrable, that is, a f (t)dt < ∞ for any
finite interval a < x < b. Then the distribution T f : D → R associated to the function f (t) is defined as
T f (φ) =
where φ is any test function in D.
Z
∞
fφ dt
−∞
Example 2.56. Let fn be the function defined in (2.64). Then
Z ∞
fnφ dt
Tfn(φ) =
−∞
=n
Z t0 + 1
n
φdt.
t0
Let Tn n = 1, 2, 3, be a sequence of distributions. Then notice for a fixed test function φ, Tn(φ) is
a sequence of real numbers. We now describe the limit of a sequence of distributions.
Definition 2.57. The sequence of distributions Tn n = 1, 2, 3, has limit equal to the distribution
T0 only if for each test function φ the sequence of real numbers Tn(φ) has limit equal to T0(φ).
In other words Tn → T0 as distributions only if Tn(φ) → T0(φ) as real numbers for each test function
φ. The following lemma will be useful in proving that the ‘pulse’ functions fn defined in (2.64) has
limit equal to the Dirac delta function δ(t − t0) in the sense of distributions, that is, Tfn → δ(t − t0)
as n → ∞.
Lemma 2.58. Let φ(t) be a continuous function and let fn be defined as in equation (2.64) above.
Then
Z ∞
fn(t)φ(t)dt = φ(t0).
lim
n→∞
Proof. From equation (2.64)
Z
∞
fn(t)φ(t)dt = n
−∞
Z t0 + 1
n
−∞
φ(t)dt.
t0
=n
Z t0 + 1
n
(φ(t) − φ(t0) + φ(t0) )dt
Z
(φ(t) − φ(t0))dt + n
t0
=n
1
t0 + n
t0
Z t0 + 1
n
t0
(2.65)
φ(t0)dt
101
2.3 Applications and more properties of Laplace transforms
Let us consider the first integral on the right. As φ(t) is continuous at t = t0, by the definition of
continuity, given any ǫ > 0 there exists a δ > 0 such that
− ǫ < (φ(t) − φ(t0)) < ǫ
when
1
n
− δ < t − t0 < δ.
(2.66)
Now as δ > 0, there exists an integer N such that < δ for all n > N . Integrating the first inequality
1
in (2.66) between t0 + n and t0 for any n > N gives
Z t0 + 1
n
t0
( − ǫ)dt <
Z t0 + 1
n
(φ(t) − φ(t0))dt <
t0
Z t0 + 1
n
(2.67)
ǫdt
t0
and as ǫ is independent of t, it is easy to evaluate the first and third inequalities in (2.67)
ǫ
− <
n
Z t0 + 1
n
t0
(φ(t) − φ(t0))dt <
ǫ
n
and multiplying throughout by n we have
Z t0 + 1
n
(φ(t) − φ(t0))dt < ǫ.
−ǫ<n
(2.68)
t0
We have shown that for any ǫ > 0 there exists some N such that for all n > N the inequality (2.68)
holds. As this is true for any ǫ > 0, we can make ǫ in (2.68) as small as we wish, this therefore
implies that


Z t0 + 1
n
lim  n
(φ(t) − φ(t0))dt = 0.
(2.69)
n→∞
t0
Now consider the second integral on the right in the last line of (2.65). As φ(t0) is a constant
Z t0 + 1
n
n
t0
and therefore
1
t0 +
φ(t0)dt = nφ(t0) [ t] t0 n = φ(t0)
Z t0 + 1
Z t0 + 1
n
n
fn(t)φ(t)dt = n
(φ(t) − φ(t0))dt + n
φ(t0)dt
−∞
 t0
 t0
Z ∞
Z t0 + 1
n

(φ(t) − φ(t0))dt + φ(t0)
⇒
fn(t)φ(t)dt = n
Z
∞
−∞
t0
Taking the limit as n → ∞ gives
lim
n→∞
From (2.69) we have
Z
∞
−∞

fn(t)φ(t)dt = lim  n
n→∞
lim
n→∞
Z
∞
Z t0 + 1
n
t0
(φ(t) − φ(t0))dt + lim φ(t0)
n→∞
fn(t)φ(t)dt = 0 + lim φ(t0)
n→∞
−∞
and as φ(t0) is a constant we get our desired result
Z ∞
fn(t)φ(t)dt = φ(t0).
lim
n→∞

−∞
We now use Lemma 2.58 to show that the ‘pulse’ functions fn defined in (2.64) has limit equal to
the Dirac delta function δ(t − t0) in the sense of distributions.
Lemma 2.59. Let fn be defined as in equation (2.64) and let Tfn be the corresponding distributions as in Definition 2.55. Then as n → ∞
Tfn → δ(t − t0)
102
Laplace transforms
as distributions.
Proof. By Definition 2.57 Tfn → δ(t − t0) as n → ∞ if for each test function φ,
Now from Definition 2.55
lim Tfn(φ) = δ(t − t0)(φ)
n→∞
Tfn(φ) =
Z
∞
(2.70)
fnφ dt
−∞
and from the definition of the Dirac delta as a distribution in Definition 2.54
δ(t − t0)(φ) = φ(t0).
Substituting these two definition into (2.70) we have that Tfn → δ(t − t0) as n → ∞ if for each test
function φ
Z ∞
fnφ dt = φ(t0).
(2.71)
lim
n→∞
−∞
However since any test function φ is continuous, (2.71) is true from Lemma 2.58 and therefore the
result follows.
We now determine the Laplace transform of the Dirac delta function δ(t − t0). Recall that δ(t − t0) is
a distribution. The following is the definition of the Laplace transform of a distribution (also see
Reinhard [3]).
Definition 2.60.
Let
φs =
0
when t < 0
e−st when t > 0
If T is a distribution and if T (φs) exists then T has Laplace transform
L[T ] = T (φs)
where T (φs) is the evaluation of the functional T at the function φs.
Using this definition we have
Theorem 2.61. The Laplace transform of the Dirac delta function δ(t − t0) where t 0 > 0 is
L[δ(t − t0)] = e−st0.
(2.72)
Proof. As stated in Definition 2.54 the Dirac delta is a distribution. As a functional δ(t − t0) takes
any function φ and evaluates that function at t = t0, hence the functional δ(t − t0) takes the function
φs and evaluates it at t = t0, that is
as t0 > 0. By Definition 2.60 we have
δ(t − t0)(φs) = e−st0
L[δ(t − t0)] = δ(t − t0)(φs) = e−s t0.
By letting t0 = 0 in the formula (2.72) we have the following
Corollary 2.62. The Laplace transform of the Dirac delta function δ(t) is
L[δ(t)] = 1.
2.3 Applications and more properties of Laplace transforms
103
In this course we shall treat δ(t − t0) as a function. The above precise
definition of δ(t − t0) is only for the interest of the reader, it is not
required material. Knowledge of the formulae
L[δ(t − t0)] = e−st0
L[δ(t)] = 1
is required as is the application of these formulae as done in the following example.
Example 2.63. Solve the differential equation
y ′′ + 9y = δ(t − 1)
(2.73)
subject to the initial conditions y(0) = 0, y ′(0) = 0.
Answer: Take the Laplace transform of both sides of the given differential equation
⇒
L[y ′′ + 9y] = L[δ(t − 1)]
L[y ′′] + 9L[y] = e−s
Using formula (2.45) for the transform of the second derivative we have
s2L[y(t)] − sy(0) − y ′(0) + 9L[y(t)] = e−s
Substituting the given initial conditions y(0) = 0, y ′(0) = 0 gives
s2L[y(t)] + 9L[y(t)] = e−s
Let L[y(t)] = Y (s)
s2Y (s) + 9Y (s) = e−s
Now solve for Y (s)
(s2 + 9)Y (s) = e−s
e−s
3
1
Y (s) = 2
= e−s
(s + 9)
3 (s2 + 32)
3
3
1
1
Notice that e−s 3 (s2 + 32) is of the form e−st0Y (s) with e−st0 = e−s and Y (s) = 3 (s2 + 32) . Now
1
3
1
Y (s) =
⇒ y(t) = L−1[Y (s)] = sin(3t)
3 (s2 + 32)
3
and by the ‘t-shift’ formula of Corollary 2.34
3
1
L−1 e−s
= H(t − 1)y(t − 1)
3 (s2 + 32)
1
= H(t − 1) sin(3(t − 1))
3
and so the solution to this initial value problem is
1
y(t) = H(t − 1) sin(3(t − 1)).
3
One can associate a physical interpretation to Example 2.63. The homogeneous part of the differrential equation (2.73), that is
y ′′ + 9y = 0
is an equation that describes some type of oscillation about an equilibrium position, for example the
simple harmonic motion of a mass loaded on a spring, where y measures the displacement from the
equilibrium position
104
Laplace transforms
y
y = 0(equilibrium position)
The inhomegeneous part of the differential equation
y ′′ + 9y = δ(t − 1)
(2.74)
that is, the right hand side of the equation (2.74), may be interpreted in our example of a spring as
a vertical outside force applied to the mass over time. In this case δ(t − 1) can be viewed as a
sudden impulse at an instant of time when t = 1.
Now the initial condition y(0) = 0 means that the mass is at the equilibrium position at time t =
0. The condition y ′(0) = 0 means that the velocity at time t = 0 is zero. The mass is therefore at rest
and does NOT move until time t = 1 when the impulse δ(t − 1) is applied, causing the mass to oscillate as we can see from the following graph of the solution
1
y(t) = H(t − 1) sin(3(t − 1))
3
(2.75)
of the differential equation (2.74)
y(t)
1/3
t
1
Also notice from the graph of y(t) that after time t = 1 the system oscillates freely about the equilibrium position as there is no outside force after t = 1.
105
2.3 Applications and more properties of Laplace transforms
2.3.5 Differentiation of transforms
The result given in the following is useful for determining tranforms and inverse transforms.
Theorem 2.64. If Y (s) = L[y(t)] then Y ′(s) = L[ − t y(t)] where Y ′(s) denotes the derivative of
Y (s) with respect to the variable s.
Proof. From the definition of the Laplace transform
Z ∞
Y (s) =
e−sty(t)dt
0
and differentiating both sides of this equation with respect to s gives
Z ∞
d
e−sty(t)dt.
Y ′(s) =
ds 0
It is possible to
2.7exchange
(2.76)
the differentiation and integration operations in equation (2.76)
Z ∞
d −st
Y ′(s) =
( e y(t)) dt
ds
0
and differentiating e−sty(t) partially with respect to s we have
Z ∞
′
Y (s) =
− te−sty(t) dt
Z0 ∞
=
e−s t − ty(t) dt
(
0
)
= L[ − ty(t)].
The following two examples are applications of Theorem 2.64.
Example 2.66. Find L[t sin t].
Answer:
1
s2 + 1
′
1
L[ − t sin t] = 2
s +1
− 2s
= 2
(s + 1)2
L[sin t] =
and by Theorem 2.64
and therefore by linearity
L[t sin t] =
2s
.
(s2 + 1)2
Using Theorem 2.64 gives a more efficient method of proving L[tn] =
n!
than Example 2.7.
sn+1
Example 2.67. Use the principle of induction, the definition of the Laplace transform and Theorem
2.64 to show that
n!
L[tn] = n+1
s
when n = 0, 1, 2, 2.7. This is done by an application of the following
∂y(s, t)
Lemma 2.65. Let y(s, t) and
be continuous on the set {(s, t)|c 6 s 6 d, 0 6 t < ∞}; let the integral
∂s
R ∞
be uniformly convergent and denote Y (s) = 0 y(s, t) dt. Then Y (s) is differentiable and
Z ∞
∂y(s, t)
Y ′(s) =
dt.
∂s
0
Knowledge of this lemma is not required for this course.
R
∞ ∂y(s, t)
0
∂s
dt
106
Laplace transforms
Answer: The statement is true when n = 0 as
L[t0] = Z
L[1]
∞
e−stdt
Z R
= lim
e−stdt
R→∞ 0
−st R
e
= lim
R→∞ − s 0
1 e−sR
= lim
−
s
s
R→∞
1
when s > 0
=
s
0!
= 0+1
s
=
0
Assume the statement is true when n = k, that is
L[tk] =
Then by using Theorem 2.64
L[ − t.tk] =
k!
sk+1
.
k!
′
sk+1
′
1
⇒ L[ − tk+1] = k! k+1
s
− (k + 1)
= k!
sk+2
− (k + 1)!
=
sk+2
and therefore
L[tk+1] =
(k + 1)!
sk+2
and so the statement is true when n = k + 1, hence by the principle of induction
L[tn] =
is true for all n = 0, 1, 2, n!
sn+1
2.3.6 The Gamma function Γ(x)
In Examples 2.7 and 2.67 we obtained the formula
L[tn] =
n!
sn+1
when n = 0, 1, 2, . We now wish to obtain a more general formula, that is, a formula for L[tx]
when x is a real number and x > − 1. To do this, we need to define the gamma function Γ(x).
Definition 2.68. The gamma function Γ(x) is defined when x > 0 as an integral
Z ∞
Γ(x) =
e−ttx−1 dt.
0
The following result gives some properties of the gamma function.
Lemma 2.69. Let Γ(x) denote the gamma function. Then
i. Γ(1) = 1
ii. Γ(x + 1) = x Γ(x) when x > 0
iii. Γ(n + 1) = n!
when n = 0, 1, 2, (2.77)
107
2.3 Applications and more properties of Laplace transforms
√
1
iv. Γ 2 = π
Proof.
i. Let x = 1 in (2.77)
Γ(1) =
∞
Z
e−tt1−1 dt.
Z0 ∞
e−tdt
Z R
e−tdt
= lim
=
0
R→∞
0
= lim [ − e−t]R
0
R→∞
= lim 1 − e−R
R→∞
=1
ii. Replacing x by x + 1 in (2.77)
Z
∞
e−ttx dt
Z R
e−ttxdt
= lim
Γ(x + 1) =
0
R→∞
(2.78)
0
now use integration by parts with
v = tx
dv = xtx−1 dt
to get
Γ(x + 1) = lim
R→∞
Z
R
du = e−t dt
u = − e−t
e−ttxdt
0
∞
e−ttx−1 dt
R→∞
0
Z R
e−ttxdt
= lim e0.0x − e−RRx + x lim
R→∞ 0
R→∞
= lim − e−RRx + x Γ(x)
= lim
[ − e−ttx]R
0 +x
Z
R→∞
and the limit
− Rx
R→∞ eR
lim − e−RRx = lim
R→∞
∞
is of the indeterminate form ∞ . Applying L’Hospitals rule by differentiating the numerator
and denominator ⌈x⌉ times with respect to R, where ⌈x⌉ is the the smallest integer > x, we
have that
lim − e−RRx = 0
R→∞
and therefore
Γ(x + 1) = x Γ(x)
and clearly this result holds for x > 0 as Γ(x) is defined for x > 0.
iii. We prove Γ(n + 1) = n! when n = 0, 1, 2, by the principle of induction.The statement is true
when n = 0 as
Γ(1) = 1
from part i. Assume the statement is true when n = k, that is
Γ(k + 1) = k!
Then letting x = k + 1 in
Γ(x + 1) = x Γ(x)
(2.79)
108
Laplace transforms
of part ii) we have
Γ(k + 2) = (k + 1) Γ(k + 1)
and using (2.79) we have
Γ(k + 2) = (k + 1)k!
= (k + 2)!
so the statement is true for n = k + 1, hence by the principle of induction
Γ(n + 1) = n!
is true for all n = 0, 1, 2, iv. Let x =
1
in (2.77)
2
Z ∞
1
−
1
Γ
=
e−tt 2 dt
2
0
using the substitution
1
1
1 −
du = t 2 dt
2
u = t2
in (2.80) we get
(2.80)
Z ∞
2
1
Γ
=2
e −u du
2
0
(2.81)
Using double integrals (a topic done later in this course) we can determine the right side of
(2.81)
Z ∞ Z ∞
Z ∞
Z ∞
2
2
2
2
e−y dy =
e−x e− y dxdy
e−x dx
−∞
−∞
−∞
−∞
Z ∞ Z ∞
2
2
=
e−(x + y )dxdy
−∞
Changing variables from Cartesian (x, y) to polar
Z ∞ Z ∞
Z
−(x2 +y 2)
e
dxdy =
−∞
−∞
=
2π
2π
Z
Z
2π
0
⇒
∞
−∞
∞
"
e
−
e−r
2
−r 2
rdr dθ
#
∞
2
dθ
0
1
dθ
2
=π
Z ∞
Therefore
Z
Z
0
0
0
=
−∞
(r, θ)2.8
2
e−x dx
Z ∞
−∞
2
e−y dy = π
−∞
2
2
e−x dx = π
since x, y are dummy variables and therefore
Z ∞
√
2
e−x dx = π
−∞
2
As e−x is symmetrical about the y − axis
√
Z ∞
π
−x2
e dx =
2
0
and substituting (2.82) into (2.81) we get our desired result
√
1
Γ
= π.
2
2.8. see Section 9.3 of Kreyszig [2]
(2.82)
2.3 Applications and more properties of Laplace transforms
109
Having defined the gamma function, we can now determine L[tx] when x > − 1.
Theorem 2.70.
Let x be a real number such that x > − 1. Then
Γ(x + 1) 2.9
.
sx+1
L[tx] =
Proof. Substitute x + 1 for x in the definition of Γ(x)
Z ∞
Γ(x + 1) =
e−ttx dt
0
Using the substitution
t = su
dt = sdu
where we treat s as a constant and u as the variable, we get
Z ∞
Γ(x + 1) =
e−su(su)x s du
0
Z ∞
= sx+1
e−suux du
0
and as u is a dummy variable, we can replace u by another dummy variable t to get
Z ∞
2.10
Γ(x + 1) = sx+1
e−sttx dt
0
and from the definition of the Laplace transform of tx
Γ(x + 1) = sx+1L[tx]
which gives our desired result
L[tx] =
Γ(x + 1)
.
sx+1
(2.83)
The following is an application of Theorem 2.70.
√
5
Example 2.71. Show that L t 2 =
15 π
7
.
8s 2
By Theorem 2.70
L t
and from Lemma 2.69 part ii)
5
2
=
7
Γ 2
7
s2
5
7
5
= Γ
Γ
2
2
2
and applying Lemma 2.69 part ii) two more times we have
7
5
5
Γ
= Γ
2
2
2
5 3
3
=
Γ
2 2
2
531
1
=
Γ
222
2
2.9. We require x > − 1 because in the definition of Γ(x) we have x > 0
2.10. If s < 0 then the integral diverges from Lemma 2.3. and clearly (2.83) is undefined for s = 0, so it is necessary that s >
0 for (2.83) to hold.
110
and from Lemma 2.69 part iv)
and therefore
Laplace transforms
√
15 π
5 3 1√
7
=
Γ
π=
222
2
8
√
5
15 π
L t2 =
7 .
8s 2
Chapter 3
Fourier series
3.1 Definitions
Definition 3.1. Let f (x) be a function that is defined for all x. Then f (x) is called periodic if
there exists a positive number p such that
f (x + p) = f (x)
for each x. The number p is called a period of the function f (x).
Example 3.2. The function sin x has period p = 2π as one can check
sin(x + 2π) = sin(x)
(3.1)
by using the trigonometric formula sin(x + y) = sin x cos y + cos x sin y:
sin(x + 2π) = sin(x)cos(2π) + cos(x)sin(2π)
= sin(x).(1) + cos(x).(0)
= sin(x)
Equation (3.1) can be interpreted graphically as follows. The graph of sin(x + 2π) is a shift of the
graph of sin(x) to the left by 2π units. Notice that this ‘shifted’ graph sin(x + 2π) coincides with the
original sin(x) and this implies equality in (3.1).
1
sin(x)
0
-1
period
= 2pi
Note 3.3. In the following, we shall always let the period p = 2L. We do this to be consistent with
notation used in a later section of partial differential equations. For example, in the case of Example
3.2, sin(x) has period 2L = 2π and clearly L = π.
Definition 3.4. Let the functionf (x) be a periodic function of period 2L. Then
∞
nπx nπx
a0 X +
+ bnsin
ancos
L
L
2
n=1
111
(3.2)
112
Fourier series
is a Fourier series for f (x) if its coefficients are given by the formulae
Z
nπx
1 L
dx
where n = 0, 1, 2, 3, f (x)cos
an =
L
L −L
Z
1 L
nπx
bn =
f (x)sin
dx
where n = 1, 2, 3, L −L
L
(3.3)
(3.4)
Equations (3.3) and (3.4) are called Euler’s formulae and the numbers an , bn are called the
Fourier coefficients of f (x).
Example 3.5. Let
f (x) = x where − π < x 6 π
and
f (x + 2π) = f (x)
(3.5)
Sketch the graph of f (x) and find the Fourier series of f (x) on the interval ( − π, π).
Answer: Note that the function f (x) is defined to be periodic with period 2π by the statement
f (x + 2π) = f (x)
in (3.5). The graph of f (x) is obtained by first drawing f (x) = x on the interval − π < x 6 π and
then repeatedly shifting this drawing by multiples of 2π
Figure 3.1. Graph of f (x)
Notice the closed dots of Figure 3.1 imply that f (x) = π when x = , − π, π, 3π, 5π, We need to determine the Fourier coefficients an and bn. It is typical to split the case of an into
two subcases n = 0 and n = 1, 2, this is because n = 0 causes the cosine term in (3.3) to become 1.
Z
0πx
1 L
x cos
dx
L
L Z −L
1 π
=
x cos 0dx
π Z −π
π
1
xdx
=
π −π
2 π
1 x
=
π 2 −π
=0
a0 =
as L = π
as cos 0 = 1
113
3.1 Definitions
and so
(3.6)
a0 = 0
Now determine an when n = 1, 2, Z
1 L
nπx
dx
x cos
L Z −L
L
π
1
nπx
=
x cos
dx
π Z−π
π
1 π
x cos nx dx
=
π −π
an =
notice the π ′s cancel.
Integrating by parts with
v=x
dv = 1 dx
du = cos nxdx
sin nx
u=
n
gives
π
Z
1 x sin nx
1 π sin nx
an =
−
dx
π
π −π
n
n
π−π
1 h cos nx iπ
1 x sin nx
+
=
n
n2
π
−π
π
−π
1
π sin nπ cos nπ
− π sin ( − nπ) cos ( − nπ)
+
+
=
−
n2
n2
π
n
n
recall when n is an integer
expression
value
sin nπ
0
sin ( − nπ)
0
cos nπ
( − 1)n
cos ( − nπ) ( − 1)n
Table 3.1.
and therefore
1
an =
π
which gives
( − 1)n
0+
n2
( − 1)n
− 0+
n2
an = 0 when n = 1, 2, 3, And now determine the bn when n = 1, 2, Z
1 L
nπx
dx
x sin
L Z −L
L
1 π
nπx
=
x sin
dx
π Z−π
π
1 π
x sin nx dx
=
π −π
bn =
Integrating by parts with
v=x
dv = 1 dx
du = sin nxdx
cos nx
u=−
n
(3.7)
114
Fourier series
gives
Z
1 h x cos nx iπ
cos nx
1 π
dx
−
−
−
π
n
n
−π
π −π
π
h
i
1
x cos nx π
1 sin nx
=
−
+
n2
π
n
−π
π
−π
1
π cos nπ sin nπ
− π cos ( − nπ) sin ( − nπ)
=
+
+
−
− −
π
n2
n2
n
n
bn =
and using the values from Table 3.1
1
− π ( − 1)n
π( − 1)n
bn =
+0 − −
+0
−
π
n
n
π( − 1)n π ( − 1)n
1
−
=
−
n
π
n
2( − 1)n
=−
n
which gives
bn =
2( − 1)n+1
when n = 1, 2, 3, n
(3.8)
Substituting L = π and the expressions for a0, an and bn given by (3.6),(3.7) and (3.8) into
∞
we get our answer
nπx a0 X nπx
+
+ bnsin
ancos
L
2
L
n=1
the Fourier series of the function f (x) given in (3.5) is
∞
X
2( − 1)n+1
sin nx.
n
n=1
3.2 Convergence of Fourier Series
In Example 3.5 we proved that the Fourier series of
f (x) = x where − π < x 6 π
and
f (x + 2π) = f (x)
(3.9)
is
∞
X
2( − 1)n+1
sin nx.
n
(3.10)
n=1
We will see, from Theorem 3.8 and Example 3.9 below that the function f (x) given in (3.9) converges to its Fourier series (3.10) for all values of x except when x = , − π, π, 3π, 5π, Definition 3.6.
Suppose
∞
nπx nπx
a0 X +
+ bnsin
ancos
L
L
2
n=1
(3.11)
115
3.2 Convergence of Fourier Series
is the Fourier series of a function g(x). Let
m
a0 X nπx nπx
+
+ bnsin
ancos
2
L
L
(3.12)
n=1
be the sum of the first m terms of (3.11). Then the Fourier series (3.11) is said to converge to g(x)
at x = x0 if
!
m
a0 X nπx0 nπx0
lim
+
+ bnsin
= g(x0).
ancos
2
L
L
m→∞
n=1
Intuitively Definition 3.6 means that the sum of the first m terms (3.12) of the Fouries series is
an ‘approximation’ of the function g(x), and this approximation gets better as m gets larger. Let us
return to our example of f (x) given in (3.9), in which the Fourier series (3.10) converges to f (x)
except when x = , − π, π, 3π, 5π, Consider when m = 5, that is the sum of the first five terms of (3.10)
5
X
2( − 1)n+1
sin nx
n
(3.13)
n=1
and consider the graph of this function obtained using the computer program Maple
5.0
2.5
0.0
−5.0
−2.5
0.0
2.5
5.0
−2.5
7.5
10.0
x
−5.0
Figure 3.2. Graph of the first five terms of Fourier series (3.10)
Notice that this graph is approximately the same as the graph of f (x) given in Figure 3.1, hence the
function f (x) = x is ‘approximately’ equal to (3.13), that is
f (x) ≃
5
X
2( − 1)n+1
sin nx.
n
n=1
Now consider when m = 25, that is the sum of the first twenty-five terms of (3.10)
25
X
2( − 1)n+1
sin nx,
n
n=1
and the graph of the first twenty-five terms is
(3.14)
116
Fourier series
5.0
2.5
0.0
−5.0
−2.5
0.0
2.5
5.0
7.5
10.0
x
−2.5
−5.0
Figure 3.3. Graph of the first twenty-five terms of Fourier series (3.10)
and we see that the graph Figure 3.3 more closely resembles the graph of f (x) given in Figure 3.1.
When m = 1000, the graph of the first thousand terms of (3.10) is
5.0
2.5
0.0
−5.0
−2.5
0.0
2.5
−2.5
5.0
7.5
10.0
x
−5.0
Figure 3.4. Graph of the first thousand terms of Fourier series (3.10)
which is almost identical to the graph of f (x) given in Figure 3.1, except for the presence of what
appear to be vertical lines at x = − π, π, 3π, . These lines are in fact not vertical but are almost
vertical with large negative slope3.1. These almost vertical lines are present in Figure 3.4 because
1000
X
2( − 1)n+1
sin nx
n
n=1
3.1. This slope can be determined. Recall we are graphing
1000
X
n=1
2( − 1)n+1
sin nx
n
(3.15)
To get the slope of (3.15) at x = π, simply differentiate (3.15) w.r.t. x and substitute x = π; this will give
1000
X
n=1
2( − 1)n+1 cos nπ =
1000
X
n=1
2( − 1)n+1 ( − 1)n =
1000
X
n=1
2( − 1)2n+1 = − 2000
117
3.2 Convergence of Fourier Series
is a finite sum of continuous functions sin n x, therefore is itself continuous which means that its
graph cannot have gaps. These almost vertical lines are therefore necessary to maintain continuity.
It is very interesting that when m → ∞ these almost vertical lines disappear and the graph of the
Fourier series, in this case, becomes discontinuous as we see next.
It is not possible to directly plot the case of m = ∞ via computer. However from Theorem 3.8 we
can predict what the graph of m = ∞, that is the graph of the Fourier series
is
∞
X
2( − 1)n+1
sin nx.
n
n=1
Figure 3.5. Graph of the Fourier series (3.10)
and notice that this graph is identical to the graph of f (x) given in Figure 3.1 except when x = − π,
π, 3π, , that is except at the points at which our original function f (x) is discontinuous. By
examining the closed dots of Figures 3.1 and 3.5, we see that f (x) takes value π and the Fourier
series (3.10) takes value 0 at these points of discontinuity. We now explain some terms necessary for
the statement of Theorem 3.8. We do not define these terms precisely, but illustrate their meaning
via an example; see Kreyszig [2] for precise definitions.
Definition 3.7. Consider the following piecewise defined function

 x2
when x < 1
h(x) = x

when x > 1.
2
The graph of this function is
h(x)
1
1/2
x
1
118
Fourier series
Then the left hand limit at x = 1 is
lim h(x) = 1
x→1−
and is the value h(x) approaches as x tends to 1 from the left; one may think of a small object
moving along the parabola x2 heading right3.2 to just before the point (1, 1); the height of the object
is the value of limx→1− h(x).
The right hand limit at x = 1 is
lim h(x) =
x→1+
1
2
and is the value h(x) approaches as x tends to 1 from the right; again one may think of a small
1
x
object sliding left along the line to just before the point (1, ); the height of the object is the value
2
2
of limx→1+ h(x).
The left hand derivative at x = 1 is defined as
h(1 + ǫ) − (limx→1− h(x))
ǫ
(1 + ǫ)2 − 1
= lim
ǫ
ǫ→0−
= lim ǫ + 2
lim
ǫ→0−
ǫ→0−
=2
and this can be interpreted as the slope of the function h(x) at a point ‘just before’ (1, 1); in this
case this is the same as the slope of the parabola at (1, 1).
The right hand derivative at x = 1 is defined as
h(1 + ǫ) − (limx→1+ h(x))
ǫ
1
1
(1
+
ǫ)
−
2
= lim 2
ǫ
ǫ→0+
1
= lim
ǫ→0+ 2
1
=
2
lim
ǫ→0+
1
and this can be interpreted as the slope of the the function h(x) at a point ‘just after’ (1, 2 ); in this
1
case this is the same as the slope of the line at (1, 2 ).
Recall that it is not possible for a function to have a derivative at a point of discontinuity3.3 such as
x = 1 in the example above. Note however that the definitions of left hand/right hand derivative are
constructed to extend the notion of a derivative to points of discontinuity, one simply examines the
slope of the curve just before the point of disconuity to get the left hand derivative (if it exists), likewise examining the slope just after the point of discontinuity gives the right hand derivative.
We can now state a theorem that describes, for which values of x, a Fourier series of a function f (x)
is equal3.4 to the function f (x).
Theorem 3.8. Let f (x) be a function that is periodic with period 2L and that is piecewise continuous on the interval − L < x < L. Also assume that for each x that either f (x) is differentiable at x
or the left hand derivatives and right hand derivatives of f (x) exist. Let
∞
nπx nπx
a0 X +
+ bnsin
ancos
L
L
2
n=1
3.2. heading right=approach from the left
3.3. because differentiablity implies continuity
3.4. more precisely, for which values of x the Fourier series of f (x) converges to its function f (x)
(3.16)
119
3.2 Convergence of Fourier Series
be the Fourier series of f (x). Then the series (3.16) converges to f (x) at each point x except possibly at points where f (x) is discontinuous. At any such point x0 of discontinuity, the Fourier series
converges to
1
lim− f (x) + lim+ f (x) .
2 x→x0
x→x0
We now apply Theorem 3.8 to the answers in Example 3.5:
Example 3.9. Show that the Fourier series
∞
X
2( − 1)n+1
sin nx
n
(3.17)
n=1
of the function
converges to function
g(x) =
f (x) = x where − π < x 6 π
when − π < x < π
when x = π
x
0
and
and
f (x + 2π) = f (x)
g(x + 2π) = g(x).
(3.18)
Answer: We first show that f (x) satisfies the hypotheses of Theorem 3.8. By the definition of f (x)
we have
f (x + 2π) = f (x)
and so f (x) is periodic with period 2π. From the graph of f (x)
Figure 3.6. Graph of f (x)
it is clear f ′(x) = 1 and is hence is differentiable at each point x , − π, π, 3π, 5π, . At the
points of discontinuity, that is for x = (2k + 1)π where k is an integer, the left hand derivative exists:
f ((2k + 1)π + ǫ) − limx→(2k+1)π − f (x))
ǫ
ǫ→0−
π+ǫ−π
= lim
ǫ
ǫ→0−
= lim 1
lim
ǫ→0−
=1
120
Fourier series
and is equal to 1. Alternatively one can clearly see that the slope of the curve ‘just before’ the point
x = (2k + 1)π is equal to 1. Similarly the right hand derivative exists and is equal to 1 at each point
of discontinuity x = (2k + 1)π. Hence we may apply Theorem 3.8 and let g(x) be the function that
the Fourier series (3.17) converges to. Then according to Theorem 3.8, g(x) is equal to f (x) everywhere f (x) is continuous and so
g(x) = x when − π < x < π
g(x + 2π) = g(x)
(3.19)
At the point of discontinuity x = π according to Theorem 3.8
1
g(x) =
lim f (x) + lim f (x)
2 x→π −
x→π +
1
= (π + ( − π))
2
=0
and this ‘repeats’ every 2π so
g(π) = 0
g(x + 2π) = g(x)
(3.20)
and combining (3.19) and (3.20) we get our desired answer (3.18).
Therefore the Fourier series (3.17) converges to (in other words, is equal to) g(x). The graph of
g(x) is
Figure 3.7. Graph of g(x)
and notice this almost identical to the graph of f (x) in Figure 3.6.
Example 3.9 illustrates that a Fourier series of a function f can
be identical to the function f for almost all values of x.
121
3.3 Even and odd functions
3.3 Even and odd functions
If a function is even or odd we can save some work when computing its Fourier series.
Definition 3.10.
i. A function f (x) is said to be even if f ( − x) = f (x) for all values of x.
ii. A function f (x) is said to be odd if f ( − x) = − f (x) for all values of x.
Example 3.11. Determine if the following functions are even, odd or neither.
i. f (x) = x3
ii. g(x) = x4
iii. h(x) = x3 + x4
iv. k(x) = cos (2x) + 5x2 − 11x4 + |x|
Answer:
i. Replacing the x in f (x) = x3 by − x we have
f ( − x) = ( − x)3
which implies
f ( − x) = − x3
(3.21)
Comparing the right hand side of (3.21) to the definition of the original function f (x) = x3 we
see that
f ( − x) = − f (x)
3
so f (x) = x is an odd function.
ii. Replace the x in g(x) = x4 by − x to get
g( − x) = ( − x)4
which implies
g( − x) = x4
(3.22)
Comparing the right hand side of (3.22) to the definition of the original function g(x) = x4 we
see that
g( − x) = g(x)
4
so g(x) = x is an even function.
iii. Replace the x in h(x) = x3 + x4 by − x to get
which implies
In this case, notice that
h( − x) = ( − x)3 + ( − x)4
h( − x) = − x3 + x4
h(x) = x3 + x4
and
(3.23)
− h(x) = − x3 − x4
and that the right hand side of (3.23) is equal to neither of these. Therefore h(x) is neither
even nor odd.
122
Fourier series
iv. Replace the x in
by − x to get
k(x) = cos (2x) + 5x2 − 11x4 + |x|
k( − x) = cos( − 2x) + 5( − x)2 − 11( − x)4 + | − x|
= cos(2x) + 5x2 − 11x4 + |x|
= k(x)
and so k(x) is an even function.
It is slightly more difficult to determine if a periodic function that is piecewise defined is even or
odd. To determine if such a function, for example
x
when − π < x < π
and g(x + 2π) = g(x).
g(x) =
0
when x = π
is even or odd, one may check the function on one period as the following lemma states.
Lemma 3.12.
i. A periodic function of period 2L is even if f ( − x) = f (x) on the interval − L 6 x 6 L.
ii. A periodic function of period 2L is odd if f ( − x) = − f (x) on the interval − L 6 x 6 L
Proof. Not required.
Example 3.13. Show that the function
x
when − π < x < π
g(x) =
0
when x = π
is odd.
and
g(x + 2π) = g(x)
Answer: For any value of x in the interval − π < x < π we have
g( − x) = − x = − g(x)
and when x = π or − π
g( − x) = 0 = − g(x)
and hence
g( − x) = − g(x)
on the interval − π 6 x 6 π and it follows from Lemma 3.12 that g(x) is odd. (the graph of g(x) is
given in Figure 3.7).
Lemma 3.14.
Let f (x) be an odd function. Then for any L > 0
Z L
f (x)dx = 0.
−L
Proof. Using a property of definite integrals
Z L
Z 0
Z
f (x)dx =
f (x)dx +
−L
−L
0
L
f (x)dx
(3.24)
123
3.3 Even and odd functions
From a change of variables u = − x on the first integral on the right of (3.24) we have
Z 0
Z 0
f (x)dx =
f ( − u)( − du)
−L
L
Z 0
=
− f (u)( − du)
Z L0
=
f (u)du
L
Z L
=−
f (u)du
Z0 L
=−
f (x)dx
and using this in (3.24) we have
Z L
−L
0
f (x)dx = −
Z
0
L
f (x)dx +
Z
L
f (x)dx = 0.
0
A more intuitive reason for Lemma 3.14 follows from the graph of an odd function f (x)
R 0
R L
f (x)dx. The area to the left represents −L f (x)dx
in which the total shaded area represents
−L
and
R L is equal in magnitude but opposite in sign to the area on the right which represents
f (x)dx; clearly these two cancel each other in equation (3.24).
0
Lemma 3.15. The Fourier series of an even function f (x) has no sine terms, that is, bn = 0 for each
n = 1, 2, 3, .
Proof. Let
∞
nπx nπx
a0 X +
+ bnsin
ancos
L
L
2
n=1
be the Fourier series of the even function f (x). From Euler’s formulae
Z
1 L
nπx
bn =
dx
where n = 1, 2, 3, f (x)sin
L −L
L
nπx
As f (x) is even and sin L
is odd
nπ( − x)
nπx f ( − x)sin
= f (x) − sin
L
L
nπx
= − f (x)sin
L
nπx
therefore f (x)sin L is an odd function. From Lemma 3.14
Z L
nπx
f (x)sin
dx = 0
L
−L
124
Fourier series
and clearly this implies bn = 0 for each n = 1, 2, 3, .
So we have
The Fourier series of an even function takes the form
∞
a0 X
nπx
+
ancos
2
L
n=1
Similarly, we can prove
Lemma 3.16.
The Fourier series of an odd function f (x) has no cosine terms, that is, an = 0 for each n = 0, 1, 2,
3, .
Therefore we also have
The Fourier series of an odd function takes the form
∞
X
bnsin
n=1
nπx
L
Example 3.17. Let
f (x) = 1 − x2 where − 1 < x 6 1
and
f (x + 2) = f (x)
(3.25)
Sketch the graph of f (x) and find the Fourier series of f (x).
Answer: The graph of f (x) is obtained by first drawing f (x) = 1 − x2 on the interval − 1 < x 6 1
and then repeatedly shifting this drawing by multiples of 2.
f (x) is clearly periodic with period 2L = 2. Notice that
f ( − x) = 1 − ( − x)2
= 1 − x2
= f (x)
when − 1 6 x 6 1 and from Lemma 3.12 f (x) is even. Therefore from Lemma 3.15, the Fourier series
of f (x) takes the form
∞
nπx
a0 X
+
ancos
(3.26)
2
L
n=1
125
3.3 Even and odd functions
that is, bn = 0 when n = 1, 2, 3, . We use Euler’s formula to determine an. Split the case of an into
two subcases n = 0 and n = 1, 2, Z
1 L
0πx
a0 =
dx
(1 − x2) cos
L −L
L
Z 1
1
=
(1 − x2)dx
as L = 1
1 −1
1
x3
= x−
3 −1
4
=
3
Now determine the an when n = 1, 2, Z
1 L
nπx
dx
(1 − x2) cos
L −L
L
Z
1 1
nπx
=
dx
(1 − x2) cos
1
1
Z −1
an =
π
=
−π
Integrating by parts with
(1 − x2) cos nπx dx
v = 1 − x2
gives
du = cos nπxdx
sin nπx
dv = − 2x dx
u=
nπ
1
Z 1
2
2x sin nπx
(1 − x ) sin nπx
dx
−
−
an =
nπ
nπ
−1
−1
Z 1
2x sin nπx
=0+
dx
nπ
−1
Integrating by parts again with
v = 2x
sin nπx
dx
nπ
cos nπx
u=−
n 2π 2
du =
dv = 2 dx
gives
Z 1
2x sin nπx
an =
dx
nπ
1
−1
Z 1
2cos nπx
2x cos nπx
−
−
= −
dx
n 2π 2
n2π 2
−1
1−1 1
2sin nπx
2x cos nπx
+
= −
n 3π 3
n2π 2
−1 −1
2cos nπ
2( − 1) cos ( − nπ )
2sin nπ 2sin ( − nπ)
=− 2 2 − −
+ 3 3 −
nπ
n 2π 2
n3π 3
nπ
and recall when n is an integer
expression
value
sin nπ
0
sin ( − nπ)
0
cos nπ
( − 1)n
cos ( − nπ) ( − 1)n
and therefore
an =
Hence the Fourier series of f (x) is
− 4( − 1)n
n 2π 2
∞
when n = 1, 2, 2 X − 4( − 1)n
+
cos nπx.
n 2π 2
3
n=1
126
Fourier series
3.4 Half range expansions
Let a function f (x) be defined on an interval 0 6 x 6 L. We want to be able to represent f (x) by
two types of Fourier series – a Fourier series that consists of cosine terms only or a Fourier series
that consists of sine terms only.3.5 Essentially, this may be done because we only want the Fourier
series to represent the function for half of the period − L 6 x 6 L.
Example 3.18. Let f (x) = x on the interval 0 6 x 6 π. Then we will show in Example 3.24 that we
can represent this function f (x) by a Fourier series that consists of cosine terms only, which in this
case is
∞ π 2 X ( − 1)n − 1
x= +
cos nx
n2
2 π
n=1
and this equality holds when 0 6 x 6 π. We can also represent f (x) by a Fourier series that consists
of sine terms only, which in this case is
x=2
∞
X
( − 1)n+1
sin nx
n
n=1
and this equality holds when 0 6 x < π.
Definition 3.19.
i. A Fourier series that consists of cosine terms only is called a Fourier cosine series.
ii. A Fourier series that consists of sine terms only is called a Fourier sine series.
Hence, given a function f (x) defined on an interval 0 6 x 6 L, we want to be able to represent f (x)
by either a Fourier cosine series or a Fourier sine series.
Recall from Lemma 3.15 that the Fourier series of an even function consists of cosine terms only, in
other words, the Fourier series of an even function is a Fourier cosine series. Also recall from Lemma
3.16 that the Fourier series of an odd function is a Fourier sine series. We use these facts to obtain
our desired representation of f (x) by either a Fourier cosine series or a Fourier sine series.
We obtain the representation of f (x) by a Fourier cosine series by first constructing an even
periodic function that is equal to f (x) on the interval 0 6 x 6 L. Such a function is called an even
periodic extension of f (x). Similarly we obtain the representation of f (x) by a Fourier sine series
by constructing an odd periodic function that is equal to f (x) on the interval 0 6 x 6 L. In this
case, such a function is called an odd periodic extension of f (x).
Definition 3.20. Let a function f (x) be defined on an interval 0 6 x 6 L. Then
i. the function defined as
E(x) =
f (x)
f ( − x)
06x6L
−L<x<0
E(x + 2L) = E(x)
,
is the even periodic extension of f (x) of period 2L.
ii. the function defined as
O(x) =
f (x)
− f ( − x)
06x6L
−L<x<0
is the odd periodic extension of f (x) of period 2L.
3.5. We will use this when solving partial differential equations.
,
O(x + 2L) = O(x)
127
3.4 Half range expansions
Example 3.21. Let f (x) = x on the interval 0 6 x 6 π. Then
i. the even periodic extension of f (x) is
x
06x6π
E(x) =
−x −π<x<0
,
E(x + 2π) = E(x)
and from the graph of E(x)
we see that E(x) is an even periodic function of period 2π. Also note that
E(x) = f (x)
when 0 6 x 6 π .
ii. the odd periodic extension of f (x) is
x
06x6π
O(x) =
− ( − x) − π < x < 0
,
O(x + 2π) = O(x)
and from the graph of O(x)
we see that O(x) is an even periodic function of period 2π. Also note that
O(x) = f (x)
when 0 6 x 6 π .
We now define the half-range expansions of a function f (x).
Definition 3.22. Let f (x) be defined on an interval 0 6 x 6 L.
i. The Fourier series of the even periodic extension E(x) of f (x) is called the cosine halfrange expansion of f (x) or Fourier cosine series of f (x).
ii. The Fourier series of the odd periodic extension O(x) of f (x) is called the sine half-range
expansion of f (x) or Fourier sine series of f (x).
128
Fourier series
We explain the idea of Definition 3.22. Consider the case of the cosine half-range expansion of f (x).
Recall from Theorem 3.8, that under certain conditions, the Fourier series of any periodic function
g(x) is equal to g(x). As E(x) is, by definition a periodic function, the Fourier series of the even
periodic extension E(x) is equal to E(x) on the interval − L 6 x 6 L and hence the Fourier series of
the even periodic extension E(x) is equal to f (x) on the interval 0 6 x 6 L as
E(x) = f (x) when 0 6 x 6 L.
As E(x) is even, its Fourier series consist of cosine terms only. Summarizing, the Fourier series of
the even periodic extension E(x) is a cosine series that is equal to f (x)3.6 on the interval 0 6 x 6 L.
By a similar reasoning, the Fourier series of the odd periodic extension O(x) is a sine series that is
equal to f (x) on the interval 0 6 x < L.
The following result gives formulae for the cosine and sine half-range expansions of a function.
Lemma 3.23. Let f (x) be defined on an interval 0 6 x 6 L. Then
i. the cosine half-range expansion of f (x) is given by
∞
nπx
a0 X
ancos
+
L
2
n=1
where
an =
2
L
Z
L
f (x)cos
0
where n = 0, 1, 2, 3, nπx
dx
L
ii. the sine half-range expansion of f (x) is given by
∞
X
bnsin
n=1
where
bn =
2
L
Z
L
f (x)sin
0
nπx
dx
L
nπx
L
where n = 1, 2, 3, Proof. i. By definition, the cosine half-range expansion of f (x) is the Fourier series of the even
periodic extension E(x) of f (x). As E(x) is an even function, by Lemma 3.15 the Fourier series of
E(x) consists of cosine terms only and hence is of the form
∞
nπx
a0 X
+
ancos
L
2
n=1
where
an =
f (x)
f ( − x)
for n = 0, 1, 2, 3, . But we know that
E(x) =
L
1
L
Z
E(x)cos
−L
nπx
dx
L
06x6L
−L<x<0
and so we may determine an directly from f (x):
Z
1 L
nπx
dx
an =
E(x)cos
L −L
L
!
Z L
Z 0
1
nπx
nπx
=
dx +
E(x)cos
dx
E(x)cos
L
L
L
0
−L
!
Z L
Z 0
1
nπx
nπx
=
dx +
f (x)cos
dx
f ( − x)cos
L
L
L
0
−L
3.6. provided f (x) is continuous on the interval 0 6 x 6 L, which we will assume in this class.
129
3.4 Half range expansions
Using the trigonometry identity cos( − x) = cos x we have that
Z 0
Z 0
nπx nπx
f ( − x)cos
dx =
f ( − x)cos −
dx
L
L
−L
−L
Z 0
nπu =−
f (u)cos
du
by the substitution u = − x
L
Z LL
nπu du
=
f (u)cos
L
Z0 L
nπx
=
f (x)cos
dx
as u is dummy variable
L
0
and therefore
=
1
L
2
=
L
!
Z L
nπx
nπx
f ( − x)cos
dx +
f (x)cos
dx
L
L
−L
0
!
Z L
Z L
nπx
nπx
dx +
f (x)cos
dx
f (x)cos
L
L
0
0
Z
1
an =
L
Z
0
L
f (x)cos
0
nπx
dx
L
which is our desired result.
The proof of ii) is similar to that of i).
Example 3.24. Find the
i. cosine half-range expansion
ii. sine half-range expansion
of the function f (x) = x on the interval 0 6 x 6 π.
Answer:
i. In this case L = π. From the formulae in Lemma 3.23 the cosine half-range expansion of f (x)
is given by
∞
a0 X
+
ancos nx
2
n=1
where
Z
2 π
f (x)cos nx dx
an =
π Z0
π
2
=
x cos nx dx
π 0
Split into the case of n = 0
and the case of n = 1, 2, a0 =
2
π
Z
π
x dx = π
0
Z
2 π
x cos nx dx
π0
π Z π
2
x sin nx
sin nx
=
dx
−
π
n
n
0π h 0
2
cos nx iπ
x sin nx
=
+
π
n2
0
n
0 n
1
2 ( − 1)
− 2
=
n
n2
π
an =
130
Fourier series
and therefore the cosine half-range expansion of f (x) is
∞ π 2 X ( − 1)n − 1
+
cos nx .
2 π
n2
n=1
ii. From the formulae in Lemma 3.23 the sine half-range expansion of f (x) is given by
∞
X
bnsin nx
n=1
where
Z
2 π
f (x)sin nx dx
π Z0
2 π
x sin nx dx
=
π 0
Z π
2 h x cos nx iπ
cos nx
=
dx
−
+
π
n
0
n
π 0
2 h x cos nx iπ
sin nx
=
−
+
π
n
0
n2
0
n+1
2( − 1)
=
n
bn =
and therefore the sine half-range expansion of f (x) is
2
∞
X
( − 1)n+1
sin nx .
n
n=1
Chapter 4
Partial Differential Equations
4.1 Definitions
Definition 4.1. An equation involving one or more partial derivatives of an unknown function of
two or more variables is called a partial differential equation or p.d.e. . The unknown function
is called the dependent variable. The order of the highest derivative is called the order of the
partial differential equation.
Example 4.2.
∂u
i. ∂x
=
∂ 2u
ii. ∂x2
∂u
is a first order p.d.e.
∂t
=
∂u
is a second order p.d.e.
∂t
Definition 4.3. A partial differential equation is said to be linear if it is of first degree in the
unknown function and its partial derivatives.
Example 4.4.
∂u
∂u
∂ 2u
∂u
∂ 2u
∂ 2u
i. ∂x = ∂t , ∂x2 = ∂t , ∂x2 + ∂t2 = 0 are each linear p.d.e.
2
∂u ∂u
∂ 2u
∂u
∂u
= ∂t , ∂x . ∂t = ∂x2 are each non-linear p.d.e.
ii. ∂x
iii. Let u(x, t) be a function of the two variables. Then the general form of a second order linear
p.d.e. is
a(x, t)
∂ 2u
∂ 2u
∂ 2u
∂ 2u
∂u
∂u
+
b(x,
t)
+
c(x,
t)
+
d(x,
t)
+ e(x, t)
+ f (x, t)
+ g(x, t)u = h(x, t)
∂x2
∂t2
∂x∂t
∂t∂x
∂x
∂t
where a, b, c, d, e, f , g and h are given functions of x and t.
Definition 4.5. If each term of a partial differential equation contains the dependent variable or its
partial derivatives then the equation is said to be homogeneous, otherwise it is said to be inhomogeneous.
Example 4.6.
i. ∂x2
+
∂ 2u
= 0 is a homogeneous p.d.e.
∂t2
∂ 2u
∂x2
+
∂ 2u
= x2t is an inhomogeneous p.d.e.
∂t2
∂ 2u
ii.
131
132
Partial Differential Equations
Consider a function u(x, y) of the two independent variables x and y. The domain of the function
u(x, y) will be a subset of the xy-plane. For example the real-valued function
p
u(x, y) = 1 − (x2 + y2)
has domain equal to
which is a closed disc of radius 1:
{(x, y)|x2 + y 2 6 1}
The closed disc above is an example of a subset in the space of independent variables x and
y of the function u(x, y). Note in this example, the space of independent variables x and y is simply
the xy-plane.
Definition 4.7.
A solution of a partial differential equation in a subset R of the space of independent variables
is a function for which all the partial derivatives that appear in the equation exist and satisfies the
equation everywhere in R.
Example 4.8.
∂ 2u
i. Show that the function u(x, y) = x2 − y 2 is a solution of the p.d.e. ∂x2
R = xy-plane.
∂ 2u
+
∂ 2u
= 0 in the set
∂y 2
∂ 2u
Answer: When u(x, y) = x2 − y 2, the second partial derivatives ∂x2 and ∂y 2 are
and so for u(x, y) = x2 − y 2
∂ 2u
=2
∂x2
∂ 2u
=−2
∂y 2
(4.1)
∂ 2u ∂ 2u
+
= 2 + ( − 2)
∂x2 ∂y2
=0
and notice that the second partial derivatives in (4.1) exist for all points in the xy-plane and
also these second partial derivatives satisfy the p.d.e for all points in the xy-plane. Therefore,
∂ 2u
by Definition 4.7, u(x, y) = x2 − y 2 is a solution of the p.d.e. ∂x2
y) in the xy-plane.
+
∂ 2u
= 0 at each point (x,
∂y 2
ii. Show that the function u(x, y) = ln(x2 + y 2) is a solution of the p.d.e.
∂ 2u
∂x2
+
∂ 2u
= 0 in the
∂y2
set R = {xy-plane − (0, 0)}, in other words, u(x, y) = ln(x2 + y 2) is a solution of the given
p.d.e. at all points on the xy-plane except at the origin.
133
4.1 Definitions
Answer: When u(x, y) = ln(x2 + y2), the second partial derivatives
∂ 2u 2y 2 − 2x2
=
∂x2 (x2 + y2)2
∂ 2u
∂ 2u
and 2 are
∂x2
∂y
∂ 2u 2x2 − 2y 2
=
∂y 2 (x2 + y2)2
(4.2)
and so for u(x, y) = ln(x2 + y2),
2
2
2y − 2x2
2x − 2y 2
∂ 2u ∂ 2u
+
=
+
(x2 + y 2)2
(x2 + y2)2
∂x2 ∂y 2
=0
however, notice that the second partial derivatives in (4.2) exists for all points except when
(x, y) = (0, 0), therefore by Definition 4.7, u(x, y) = ln(x2 + y 2) is a solution of the p.d.e.
∂ 2u
∂x2
+
∂ 2u
= 0 at each point (x, y) in the xy-plane except when (x, y) = (0, 0).
∂y 2
Theorem 4.9. (Superposition principle I) If u1 and u2 are solutions of a linear homogeneous
partial differential equation in some subset R in the space of independent variables then, for any
scalars c1 and c2
u = c1u1 + c2u2
is also a solution of that partial differential equation in the subset R.
Proof. (Not required).
Example 4.10. From Example 4.8 we have that
u1 = x2 − y 2
∂ 2u
are solutions of the p.d.e. ∂x2
position principle I
is a solution of
∂ 2u
∂x2
+
and
u2 = ln(x2 + y 2)
∂ 2u
= 0 in the set R = {xy-plane − (0, 0)}. Therefore by the Super∂y 2
u = 15(x2 − y 2) + 9ln(x2 + y 2)
+
∂ 2u
= 0 in the set R = {xy-plane − (0, 0)}.
∂y 2
We shall see later that we will usually obtain an infinite set of solutions u1, u2, u3, to a given linear
homogeneous p.d.e. . The following theorem states, that under certain conditions, the sum of such
an infinite set of solutions is also a solution.
Theorem 4.11. (Superposition Principle II) If each function of an infinite set u1, u2, u3, is a
solution of a linear homogeneous partial differential equation in some subset R in the space of independent variables then the function
u=
∞
X
cnun = c1u1 + c2u2 + (4.3)
n=1
is also a solution of that partial differential equation in the subset R provided the series (4.3) converges at each point in R.
Proof. (Not required).
134
Partial Differential Equations
Note 4.12. In this class, we shall use a method of solving p.d.e. that determine constants cn that
will ensure that the series
∞
X
cnun = c1u1 + c2u2 + u=
n=1
converges. This method is based on Theorem 3.8 on page 118. We will not directly test for convergence, our method will automatically guarantee convergence.
Example 4.13. One can check each of the functions
e−tcos x, e−9tcos 3x, e−25tcos 5x, , e−(2n+1) tcos(2n + 1)x, 2
is a solution of
∂u ∂ 2u
=
∂t ∂x2
in the subset R = {(x, t)|t > 0}. It is possible to prove
(4.4)
4.1that
the series
∞
X
2
1
e−(2n−1) tcos(2n − 1)x
(2n − 1)2
n=1
1
1
= e−tcos x + e−9tcos 3x + e−25tcos 5x + 25
9
(4.5)
converges at each point in the subset R = {(x, t)|t > 0}, therefore by the Superposition Principle II
the series (4.5) is a solution of the partial differential equation (4.4) in the subset R = {(x, t)|t > 0}.
We shall be interested in the following linear homogeneous partial differential equation of the second
order:
= c2 ∂x2
∂ 2u
∂t2
= c2 ∂x2
∂ 2u
∂x2
∂ 2u
∂r 2
∂ 2u
∂r 2
∂ 2u
∂u
∂t
one dimensional heat equation
∂ 2u
one dimensional wave equation
∂ 2u
+ ∂x2 = 0
1 ∂u
two dimensional Laplace’s equation
1 ∂ 2u
+ r ∂r + r2 ∂θ2 = 0
1 ∂u
1 ∂ 2u
Laplace’s equation in polar coordinates
∂ 2u
+ r ∂r + r2 ∂θ2 + ∂z 2 = 0
Laplace’s equation in cylindrical coordinates
1
∂
1 ∂
∂u
1 ∂ 2u
2 ∂u
r
+
sin
φ
+
= 0 Laplace’s equation in spherical coordinates
2
r
∂r
∂r
sin φ ∂φ
∂φ
sin2 φ ∂θ2
4.1. A direct proof of this convergence is beyond the scope of this class
135
4.2 The Heat Equation
4.2 The Heat Equation
4.2.1 A derivation of the heat equation
We illustrate how the one-dimensional heat equation
models a certain physical situation.
∂u
∂ 2u
= c2 2
∂t
∂x
(4.6)
Consider a bar of metal of length L. We wish to describe how the temperature of this bar
changes with time t. To simplify our model we assume that our rod is one-dimensional, that is, it
has no thickness. Assume also that the metal bar is insulated along the sides. We place our bar of
length L along the x-axis as follows.
Notice that as the bar is insulated along the sides, heat may escape the bar only at points
x=0
and x = L,
that is, heat can only escape at the ends of the bar.
Let u(x, t) denote the temperature of the bar at a point x and at time t. Note that the domain
of this temperature function u(x, t) is
06x6L
and 0 6 t < ∞.
We wish to model the following problem. The metal bar has an initial temperature distribution,
which is not necessarily uniform along the bar. That is, at time t = 0, the bar has a temperature
that varies along the length of the bar, so this initial temperature distribution varies with x and can
be stated as
u(x, 0) = f (x)
where 0 6 x 6 L
(4.7)
where the function f (x) describes the initial temperature distribution along the bar. The equation
(4.7) is an example of an initial condition.
For simplicity, we assume in this particular problem that the ends of the bar will be at zero temperature for all time t > 0. This may be stated as
u(0, t) = 0
for any t > 0
(4.8)
u(L, t) = 0
for any t > 0
(4.9)
Equations (4.8) and (4.9) are examples of boundary conditions.
136
Partial Differential Equations
Summarizing so far, our physical problem is as follows. We have a bar that has an
initial temperature that is given by equation (4.7). We insulate the bar along the
sides and allow the heat to escape from the ends. Therefore the temperature of the
bar will change with time. We wish to find the temperature function u(x, t) that
describes this change and will show that such a u(x, t) will be a solution of the heat
equation (4.6).
To derive the heat equation, we consider a small segment of the metal bar that lies between the
point x and x + ∆x where ∆x represents a small positive number. We represent this segment by the
grey area in the following diagram of the metal bar.4.2
As the bar is insulated along the sides, the heat flow out of the segment is only in the indicated
direction of the arrows A and B.
Recall that heat flows from one point to another point if these points are at different temperatures. Now the temperature at a point x on our metal bar at time t is given by the function
u(x, t).
Note that a difference in temperature u(x, t) from one point to the next is measured by the change
of the function u(x, t) with respect to x. By the definition of partial derivatives, such a change in u
∂u
with respect to x is given by the partial derivative ∂x . And as heat flow occurs when we have temperature change, we expect that heat flow at the point x + ∆x on the metal bar at time t should be
∂u
proportional to the partial derivative ∂x (x + ∆x, t). This is in fact true – the Law of Heat Conduction from physics tells us that the rate of heat flow at the point x + ∆x at time t (indicated by
arrow A) is given by
∂u
− k (x + ∆x, t)
(4.10)
∂x
where k is the thermal conductivity of the metal. The 8− sign ′ appears in (4.10) because heat flows
∂u
from points of high temperature to points of low temperature while ∂x is positive when temperature
increases with distance x.
Similarly the rate of heat flow at the point x at time t (as indicated by arrow B) is given by
k
∂u
(x, t).
∂x
(4.11)
Notice that the sign of (4.11) is opposite to that of (4.10) because arrow B points in the opposite
direction of arrow A. Now as heat flows out of the segment only in the direction of arrows A and B,
we have that
∂u
∂u
(4.12)
rate of heat loss in segment = k (x, t) − k (x + ∆x, t)
∂x
∂x
4.2. We draw our one dimensional metal bar as having width and height ONLY for illustrative purposes.
137
4.2 The Heat Equation
We can determine the rate of heat loss in the segment by another means. Recall that the more the
temperature of an object increases, the more heat it will contain. In fact if an object has mass m,
specific heat capacity C and temperature T (t) then, also from physics:
rate of heat increase of object = mass × specific heat capacity × rate of temperature increase
dT
= mC .
dt
Using the above physical law, it is possible to show4.3 that
rate of heat increase in segment = ρC
∂u(ξ , t)
∆x
∂t
where x 6 ξ 6 x + ∆x
where ρ is the mass per unit length of the metal bar and C is the specific heat capacity of the metal
bar. Therefore
rate of heat loss in segment = − ρC
∂u(ξ , t)
∆x
∂t
where x 6 ξ 6 x + ∆x
(4.13)
and so from equations (4.12) and (4.13) we have
− ρC
∂u(ξ, t)
∂u
∂u
∆x = k (x, t) − k (x + ∆x, t).
∂t
∂x
∂x
(4.14)
Rearranging equation (4.14) gives
∂u
∂u
(x + ∆x, t) − ∂x (x, t)
∂x
k
∂u(ξ , t)
=
ρC
∂t
∆x
!
(4.15)
and taking the limit of (4.15) as ∆x → 0 gives
k
∂u(ξ , t)
=
lim
lim
ρC
∂t
∆x→0
∆x→0
∂u
∂u
(x + ∆x, t) − ∂x (x, t)
∂x
∆x
!
.
(4.16)
Now because x 6 ξ 6 x + ∆x we have that ξ → x as ∆x → 0 and therefore
lim
∆x→0
∂u(ξ , t) ∂u(x, t)
=
.
∂t
∂t
By definition of the partial derivative
lim
∆x→0
∂u
∂u
(x + ∆x, t) − ∂x (x, t)
∂x
∆x
!
=
∂
∂x
∂ 2u(x, t)
∂u
(x, t) =
∂x2
∂x
and so from equation (4.16) we obtain
k ∂ 2u(x, t)
∂u(x, t)
=
ρC ∂x2
∂t
(4.17)
and as k, ρ and C are positive constants we may write equation (4.17) in the standard form of the
heat equation
∂ 2u
∂u
= c2 2
∂x
∂t
where c is a constant.
4.3. by using a Mean Value theorem to obtain
Z x+∆x
∂u(ξ, t)
∂u(s, t)
ds = ρC
∆x where x 6 ξ 6 x + ∆x
ρC
∂t
∂t
x
138
Partial Differential Equations
We have shown that the temperature u(x, t) of our metal bar satisifes the partial differential equation
∂u
∂ 2u
= c2 2
∂t
∂x
subject to the initial condition
u(x, 0) = f (x)
where 0 6 x 6 L
and boundary conditions
u(0, t) = 0
u(L, t) = 0
where t > 0.
In the next section we give a technique for obtaining a solution of such partial differential equations.
4.2.2 Solution of the heat equation by seperation of variables
We illustrate the use of the seperation of variables technique in the solution of the following onedimensional heat equation.
Example 4.14. Let u(x, t) denote the temperature of a thin metal bar of length 1 that is insulated
along the sides. Let the temperature along the length of the bar at time t = 0 be given by
u(x, 0) = 4x(1 − x)
where 0 6 x 6 1.
The ends of the bar are kept at a constant temperature of 0, that is
u(0, t) = 0
u(1, t) = 0
where t > 0.
Determine the temperature u(x, t) of the bar by solving the following heat equation
∂ 2u
∂u
= c2 2
∂x
∂t
(4.18)
where c is a fixed constant that depends on the physical properties of the bar.
Answer: The method of seperation of variables begins with the assumption that our solution u(x, t)
takes the form
u(x, t) = X(x)T (t),
(4.19)
that is, the function of two variables u(x, t) is ‘seperated’ into a product of a single variable function
X(x) and a single variable function T (t). Substituting the solution (4.19) into heat equation (4.18)
gives
∂2
∂
(4.20)
(X(x)T (t)) = c2 2 (X(x)T (t)).
∂x
∂t
Recall that to differentiate partially with respect to t, we treat the variable x as a constant. Therefore, when differentiating X(x)T (t) with respect to t, the function X(x) behaves as a constant and
we have
∂
(X(x)T (t)) = X(x)T ′(t)
∂t
where T ′(t) denotes the (ordinary) derivative of T (t). Similarly
∂2
(X(x)T (t)) = X ′′(x)T (t)
∂x2
139
4.2 The Heat Equation
and so from equation (4.20) we have
X(x)T ′(t) = c2X ′′(x)T (t)
⇒
T ′(t)
X ′′(x)
=
2
c T (t)
X(x)
(4.21)
X ′′(x)
T ′(t)
The left side c2T (t) of equation (4.21) is a function of the variable t; the right side X(x) of (4.21) is
a function of a different variable x. If
T ′(t)
constant
c2T (t)
T ′(t)
then the value of the function c2T (t) will vary with t. Similarly, if
X ′′(x)
constant
X(x)
X ′′(x)
then the value of the function X(x) will vary with x and, in fact, equation (4.21) will be false
T ′(t)
X ′′(x)
because we can choose values of t and x such that 2
and
take different values. The only
c T (t)
X(x)
T ′(t)
X ′′(x)
way for equation (4.21) to hold is if c2T (t) and X(x) are both constant functions and are equal to
the same constant k. So we have
T ′(t)
X ′′(x)
=
= k.
2
c T (t)
X(x)
(4.22)
We now need to determine if this constant k is zero, positive or negative. That is, we need to determine if k = 0, a2 or − a2 where a 0. Using the boundary conditions
u(0, t) = 0
u(1, t) = 0
where t > 0
(4.23)
we show that the only non-zero solutions of the heat equation (4.18) with these particular boundary
conditions 4.4occurs when k = − a2 :
k=0
If k = 0 then from equation (4.22) we obtain two ordinary differential equations
T ′(t) = 0
X ′′(x) = 0.
Solving these gives
T (t) = A ′
where A ′, B ′ and C ′ are constants.
X(x) = B ′x + C ′
(4.24)
Substituting equations (4.24) into equation (4.19) gives
u(x, t) = X(x)T (t) = Ex + F
′
′
′
(4.25)
′
where E = A B and F = A C .
From the first boundary condition of (4.23) we get
u(0, t) = E(0) + F = 0
⇒
F =0
4.4. In Example 4.15 we will see that the value k = 0 will give a non-zero solution because the boundary conditions of that
example are different from the boundary conditions of this example.
140
Partial Differential Equations
The second boundary condition gives
u(1, t) = E(1) + 0 = 0
⇒
E=0
As E = F = 0 in (4.25), we have that the solution (4.25) of our heat equation corresponding to k = 0
must be identically equal to zero.
k = a2
If k = a2 then from equation (4.22) we have
X ′′(x)
= a2
X(x)
⇒ X ′′(x) − a2X(x) = 0
⇒ X(x) = C cosh (ax) + D sinh (ax)
where C and D are constants. From equation (4.19) we have
u(x, t) = (C cosh (ax) + D sinh (ax))T (t).
(4.26)
From the first boundary condition of (4.23) we get
u(0, t) = (C cosh(0) + D sinh(0))T (t) = 0
⇒
(C(1) + D(0))T (t) ≡ 0
⇒ CT (t) ≡ 0
and this implies C = 0. The second boundary condition gives
u(1, t) = D sinh(a)T (t) ≡ 0
(4.27)
and as sinh a 0 when a 0, equation (4.27) implies that D = 0. As C = D = 0, we have that the
solution (4.26) obtained when k = a2 is identically equal to zero.
The last possibility of k = − a2 will lead to non-zero solutions of the heat equation (4.18). Subsituting k = − a2 into equation (4.22) gives
from which we obtain
X ′′(x)
T ′(t)
=
= − a2
2
X(x)
c T (t)
T ′(t) + c2a2T (t) = 0
(4.28)
X ′′(x) + a2X(x) = 0
(4.29)
and this is the point of the seperation of variables method, the assumption
u(x, t) = X(x)T (t)
converts the partial differential equation
∂ 2u
∂u
= c2 2
∂x
∂t
into two ordinary differential equations (4.28) and (4.29) which can be easily solved. By using the
D-operator method, the general solution of (4.28) is
2 2
T (t) = Ce−c a t
and the general solution of (4.29) is
X(x) = D cos(ax) + E sin(ax).
141
4.2 The Heat Equation
And so we have solutions of the heat equation (4.18) of the form
2 2
u(x, t) = e−c a t(A cos (ax) + B sin (ax))
(4.30)
where the arbitrary constants A = CD and B = CE. Note that a is also an arbitrary constant.
We now impose the boundary conditions on equation (4.30). From the first boundary condition of
(4.23) we get
2 2
u(0, t) = e−c a t(A cos(0) + B sin (0)) ≡ 0
2 2
⇒ Ae−c a t ≡ 0
⇒ A = 0.
Substituting A = 0 into (4.30) implies that our solution of the heat equation (4.18) is of the form
2 2
u(x, t) = Be−c a tsin(ax).
(4.31)
Imposing the second boundary condition of (4.23) on equation (4.31) gives
2 2
u(1, t) = Be−c a tsin(a) ≡ 0
⇒ sin(a) = 0
and so the second boundary condition forces
where k = 1, 2, 3, a = kπ
(4.32)
(note we do not consider k = 0 as this leads to the zero solution; k = − 1, − 2, do not give solutions
that are linearly independent of the solutions corresponding to k = 1, 2, ). Substituting each of
these different values of a into equation (4.31) implies that
e−c π tsin(πx), e−c (2π) tsin(2πx), e−c (3π) tsin(3πx), 2
2 2
2
2
2
are each solutions of the heat equation (4.18). So from by the Superposition Principle II on page 133
we have that
∞
X
2
2
u(x, t) =
bne−c (nπ) tsin(nπx)
(4.33)
n=1
is a solution of the heat equation (4.18) which, by construction, satisfies the boundary conditions.
Finally, we use the given initial condition
u(x, 0) = 4x(1 − x)
where 0 6 x 6 1
to determine the constants bn of equation (4.33). Setting t = 0 in (4.33) gives
u(x, 0) =
∞
X
n=1
⇒
∞
X
n=1
2
2
bne−c (nπ) 0sin(nπx) = 4x(1 − x)
bnsin(nπx) = 4x(1 − x)
when 0 6 x 6 1
when 0 6 x 6 1
(4.34)
The left side of equation (4.34) is nothing but a Fourier sine series that represents the function
4x(1 − x) on the interval 0 6 x 6 1. Recall that such a sine series is called the sine half-range
expansion of the function 4x(1 − x) on the interval 0 6 x 6 1 and from Lemma 3.23 on page 128,
the coefficients bn of the left side of (4.34) are given by
Z
nπx
2 L
f (x)sin
bn =
dx
L
L 0
Z 1
⇒ bn = 2
4x(1 − x)sin (nπx) dx
0
142
Partial Differential Equations
Integrating by parts with
v = 4x(1 − x)
gives
du = sin(nπx)dx
cos (nπx)
u=−
nπ
dv = (4 − 8x) dx
Z 1
(4 − 8x)cos (nπx)
4x(1 − x)cos (nπx) 1
+2
dx
bn = 2 −
nπ
nπ
0
0
Z 1
(4 − 8x)cos (nπx)
=0+2
dx
nπ
0
Integrating by parts again with
cos (nπx)
dx
nπ
sin (nπx)
u=
n 2π 2
v = (4 − 8x)
gives
du =
dv = − 8 dx
1
Z 1
8sin (nπx)
(4 − 8x)sin (nπx)
dx
+2
bn = 2
n 2π 2
n 2π 2
0
10
8cos (nπx)
=0+2 −
n 3π 3
0
8( − 1)n
8
=2 −
+
n 3π 3
n3π 3


when n is even
 0
=
32

 3 3 when n is odd
nπ
Substituting these values for bn into (4.33) and writing the odd values of n as 2k + 1 gives our solution
∞
X
2
2
32
e−c ((2k+1)π) tsin((2k + 1)πx)
u(x, t) =
(4.35)
3
3
(2k + 1) π
k=0
that satisfies the heat equation (4.18) subject to the given boundary and initial conditions.
In the above problem, the metal bar initially had a non-zero temperature distribution given by the
initial condition
u(x, 0) = 4x(1 − x)
where 0 6 x 6 1
and the ends of the bar are held at zero temperature. As heat flows from a high temperature to a
lower temperature, we expect the initial heat contained in the bar to flow out of the bar through the
ends which are held at zero temperature. Therefore the bar should eventually lose its heat and hence
each point in the bar should have zero temperature. This physical expectation actually agrees with
the solution (4.35) above. First notice that
32 2 2
32 2 2 lim 3 e−c π tsin(πx) = lim 3 e−c π t
lim sin(πx)
π
t→∞ π
t→∞
t→∞
= 0. lim sin(πx)
t→∞
=0
where
2 2
lim e−c π t = 0
t→∞
143
4.2 The Heat Equation
2 2
as the coefficient − c2π 2 of t in e−c π t is negative. Similarly the limit of each term in the solution
(4.35) is zero as each term in
∞
X
2
2
32
e−c ((2k+1)π) tsin((2k + 1)πx)
3
3
(2k + 1) π
k=0
32 −c2π2t
32 −c2(2π)2t
32 −c2(3π)2t
= 3e
sin(πx) +
e
sin(2πx) +
e
sin(3πx) + π
27π 3
125π 3
u(x, t) =
(4.36)
contains a exponential with a negative coefficient. Therefore we have
lim u(x, t) = lim
t→∞
t→∞
∞
X
2
2
32
e−c ((2k+1)π) tsin((2k + 1)πx)
(2k + 1)3π 3
k=0
32 −c2π2t
32 −c2(2π)2t
32 −c2(3π)2t
= lim
e
sin(πx)
+
e
sin(2πx)
+
e
sin(3πx)
+
3
27π 3
125π 3
t→∞ π
32 −c2(2π)2t
32 −c2(3π)2t
32 2 2
e
sin(2πx) + lim
e
sin(3πx) + = lim 3 e−c π tsin(πx) + lim
3
3
t→∞ 27π
t→∞ 125π
t→∞ π
=0+0+0+
= 0.
and note
lim u(x, t) = 0
t→∞
can be interpreted as meaning that the temperature of the bar u(x, t) eventually becomes zero.
4.2.3 The heat equation with insulated ends as boundary conditions
We now consider a similar heat conduction problem in a thin metal bar that is insulated along the
sides of the bar. Again we assume the bar has a initial non-zero temperature distribution. However,
unlike Example 4.14, we do not hold the ends of the bar at temperature zero. In this case we assume
the ends of the bar to be insulated (so that the bar is entirely insulated). Recall from equation
(4.10) on page 136, that the heat flow at point x on the bar is given by
−k
∂u
(x, t).
∂x
If the ends x = 0 and x = L of a bar of length L are insulated then there is no heat flow at these
points and therefore
∂u
∂u
and
− k (L, t) = 0,
− k (0, t) = 0
∂x
∂x
and hence the insulation of the ends of a bar of length L can be mathematically stated as
∂u
(0, t) = 0
∂x
∂u
(L, t) = 0.
∂x
(4.37)
The equations (4.37) is another example of boundary conditions.
Example 4.15. Let u(x, t) denote the temperature of a thin metal bar of length π that is insulated
along the sides. Let the temperature along the length of the bar at time t = 0 be given by
u(x, 0) = 1 −
x
π
where 0 6 x 6 π.
The ends of the bar are insulated for time t > 0 and therefore
∂u
(0, t) = 0
∂x
∂u
(π, t) = 0
∂x
where t > 0.
144
Partial Differential Equations
Determine the temperature u(x, t) of the bar by solving the following heat equation
∂u
∂ 2u
= c2 2
∂t
∂x
(4.38)
where c is a fixed constant that depends on the physical properties of the bar.
Answer: We use the method of seperation of variables and therefore assume that our solution u(x, t)
takes the form
u(x, t) = X(x)T (t).
(4.39)
Substituting the solution (4.39) into heat equation (4.38) gives
∂2
∂
(X(x)T (t)) = c2 2 (X(x)T (t))
∂x
∂t
⇒ X(x)T ′(t) = c2X ′′(x)T (t)
⇒
T ′(t)
X ′′(x)
=
2
c T (t)
X(x)
T ′(t)
(4.40)
X ′′(x)
and the only way for equation (4.40) to hold is if 2
and
are both constant functions and
c T (t)
X(x)
are equal to the same constant k. So we have
X ′′(x)
T ′(t)
=
=k
2
X(x)
c T (t)
(4.41)
where the constant k can be zero, negative or positive, that is
k = 0, − a2 or a2
where a 0. We now use the boundary conditions
∂u
(0, t) = 0
∂x
∂u
(π, t) = 0
∂x
where t > 0
(4.42)
to show that the only non-zero solutions of the heat equation (4.38) with the boundary conditions
(4.42) occur when k = 0 and k = − a2.
k = a2
If k = a2 then from equation (4.41) we have
X ′′(x)
= a2
X(x)
⇒ X ′′(x) − a2X(x) = 0
⇒ X(x) = C cosh (ax) + D sinh (ax)
where C and D are constants. From equation (4.39) we have
u(x, t) = (C cosh (ax) + D sinh (ax))T (t).
(4.43)
∂u
(x, t) = (aC sinh (ax) + aD cosh (ax))T (t)
∂x
(4.44)
which implies
From the first boundary condition of (4.23) we get
∂u
(0, t) = (aC sinh (0) + aD cosh (0))T (t)
∂x
⇒ (aC(0) + aD(1))T (t) ≡ 0
⇒ a DT (t) ≡ 0
145
4.2 The Heat Equation
and this implies D = 0. The second boundary condition gives
∂u
(π, t) = aC sinh(aπ)T (t) ≡ 0
∂x
(4.45)
and as sinh (aπ) 0 when a 0, equation (4.45) implies that C = 0. As C = D = 0, we have that the
solution (4.43) obtained when k = a2 is identically equal to zero.
k=0
If k = 0 then from equation (4.41) we obtain two ordinary differential equations
T ′(t) = 0
X ′′(x) = 0.
Solving these gives
T (t) = A ′
where A ′, B ′ and C ′ are constants.
(4.46)
X(x) = B ′x + C ′
Substituting equations (4.46) into equation (4.39) gives
u(x, t) = Ex + F
(4.47)
where E = A ′B ′ and F = A ′C ′. Taking the partial derivative gives
∂u
(x, t) = E
∂x
(4.48)
From the first boundary condition of (4.42) we get
∂u
(0, t) = E = 0.
∂x
The second boundary condition also gives
∂u
(π, t) = E = 0.
∂x
Notice that the boundary conditions (4.42) do not give any information about the constant F in
equation (4.47), so it is possible that F 0, and therefore when k = 0 the heat equation (4.38) may
have a non-zero solution of the form
u(x, t) = F .
(4.49)
k = − a2
The last possibility of k = − a2 will also lead to non-zero solutions. Subsituting k = − a2 into equation (4.41) gives
T ′(t)
X ′′(x)
=
= − a2
c2T (t)
X(x)
from which we obtain the two ordinary differential equations
T ′(t) + c2a2T (t) = 0
(4.50)
X ′′(x) + a2X(x) = 0
(4.51)
By using the D-operator method, the general solution of (4.50) is
2 2
T (t) = Ce−c a t
146
Partial Differential Equations
and the general solution of (4.51) is
X(x) = D cos(ax) + E sin(ax).
And so we have solutions of the heat equation of the form
2 2
u(x, t) = e−c a t(A cos (ax) + B sin (ax))
(4.52)
where the arbitrary constants A = C D and B = C E. Note that a is also an arbitrary constant. By
taking the partial derivative of (4.52) we get
2 2
∂u
(x, t) = e−c a t( − aA sin (ax) + aB cos (ax))
∂x
(4.53)
and we now impose the boundary conditions on equation (4.53). From the first boundary condition
of (4.42) we get
∂u
2 2
(0, t) = e−c a t( − aA sin(0) + aB cos (0)) ≡ 0
∂x
2 2
⇒ aBe−c a t ≡ 0
⇒ B = 0.
Substituting B = 0 into (4.30) implies that the solutions corresponding to k = − a2 is of the form
2 2
u(x, t) = Ae−c a tcos(ax)
⇒
2 2
∂u
(x, t) = e−c a t( − aA sin (ax))
∂x
(4.54)
Imposing the second boundary condition of (4.42) on equation (4.54) gives
∂u
2 2
(π, t) = e−c a t( − aA sin (aπ)) ≡ 0
∂x
⇒ sin(aπ) = 0
and so the second boundary condition forces
where k = 1, 2, 3, a=k
(note we do not consider k = 0 as this leads to the zero solution; k = − 1, − 2, do not give solutions
that are linearly independent of those corresponding to k = 1, 2, ). Substituting each of these different values of a into equation (4.53) implies that
e−c tcos(x), e−c 2 tcos(2x), e−c 3 tcos(3x), 2
2 2
2 2
are each solutions of the heat equation with the given boundary conditions. So from by the Superposition Principle II on page 133 we have that
u(x, t) =
∞
X
2 2
ane−c n tcos(nx)
(4.55)
n=1
is a solution of the heat equation that corresponds to k = − a2. From equation (4.49)
u(x, t) =
a0
2
(4.56)
is also a solution of the heat equation (that corresponds to k = 0) where we rename the constant F
a0
as 2 . Since both (4.55) and (4.56) are solutions, by the Superposition Principle their sum
u(x, t) =
∞
2 2
a0 X
ane−c n tcos(nx)
+
2
n=1
is a solution of the heat equation.
(4.57)
147
4.2 The Heat Equation
Finally, we use the given initial condition
u(x, 0) = 1 −
x
π
where 0 6 x 6 π
to determine the constants an of equation (4.57). Setting t = 0 in (4.57) gives
u(x, 0) =
∞
2
2
x
a0 X
+
ane−c (nπ) 0cos(n x) = 1 −
π
2
when 0 6 x 6 π
n=1
⇒
∞
a0 X
x
+
ancos(n x) = 1 −
2
π
when 0 6 x 6 π
(4.58)
n=1
The left side of equation (4.34) is nothing but a Fourier cosine series that represents the function
x
1 − π on the interval 0 6 x 6 π. Such a cosine series is called the cosine half-range expansion of
x
the function 1 − π on the interval 0 6 x 6 π and from Lemma 3.23 on page 128, the coefficients an
of the left side of (4.58) are given by
Z
2 L
nπx
an =
f (x)cos
dx where n = 0, 1, 2, L 0
L
Z
2 π
x
⇒ an =
1−
cos (nx) dx where n = 0, 1, 2, π 0
π
When n = 0 we have
Z
2 π
x
a0 =
1−
dx
π 0
π
π
2
x2
= x−
π
2π 0
=1
When n = 1, 2, then applying integration by parts to
Z
x
2 π
cos (nx) dx
1−
an =
π
π 0
with
x
v=1−
du = cos(nx)dx
π
sin (nx)
1
u=
dv = − dx
n
π
gives
Z
π
π
sin (nx)
2
x sin (nx)
2
dx
+ 2
1−
an =
n
n
π
π
π
0
π 0
2 cos (nx)
=0− 2
n2
π
0

 0
when n is even
=
4
 2 2 when n is odd
nπ
Substituting these values for an into (4.58) and writing the odd values of n as 2k + 1 gives our solution
∞
2
2
1 X
4
u(x, t) = +
e−c (2k+1) tcos((2k + 1)x)
(4.59)
2
2
2
(2k + 1) π
k=0
that satisfies the heat equation (4.38) subject to the given boundary and initial conditions.
In the previous problem, the metal bar initially had a non-zero temperature distribution given by
the initial condition
x
where 0 6 x 6 π.
u(x, 0) = 1 −
π
148
Partial Differential Equations
At time t = 0 the ends of the bar were insulated. As the bar is also insulated along the sides, this
implies that the bar is entirely insulated for time t > 0. Hence we expect no heat loss from the bar;
furthermore we expect the initial heat of the bar to eventually distribute itself evenly throughout the
length of the bar. So we expect as time t → ∞, that each point of the bar should have the same temperature. This physical expectation agrees with the solution (4.59) above, because
!
∞
4
1 X
−c2(2k+1)2t
+
e
cos((2k + 1)x)
lim u(x, t) = lim
(2k + 1)2π 2
t→∞
t→∞ 2
k=0
1
4
2
4
2
4
2
= + lim 2 e−c tcos(x) + lim 2 2 e−9c tcos(3x) + lim 2 2 e−25c tcos(5x) + 2 t→∞ π
t→∞ 3 π
t→∞ 5 π
1
= +0+0+0+
2
1
=
2
and note
lim u(x, t) =
t→∞
1
2
can be interpreted as meaning that the temperature of each point on the bar u(x, t) eventually
1
1
becomes equal to 2 . Also notice that 2 is the average of the initial temperature distribution
x
u(x, 0) = 1 −
where 0 6 x 6 π
π
as
R π
R π
x
x
1 − π dx
1 − π dx
0
0
1
=
=
.
length of bar
π
2
4.2.4 The heat equation with nonhomogeneous boundary conditions
The boundary conditions of the Example 4.14
u(0, t) = 0
u(1, t) = 0
where t > 0
and the boundary conditions of Example 4.15
∂u
(0, t) = 0
∂x
∂u
(π, t) = 0
∂x
where t > 0
are examples of homogeneous boundary conditions.
Definition 4.16. If the boundary conditions of the one-dimensional heat equation
can be written in the form
∂ 2u
∂u
= c2 2
∂x
∂t
where 0 < x < a and t > 0
∂u
(0, t) = 0
∂x
∂u
γu(a, t) + δ (a, t) = 0
∂x
αu(0, t) + β
where t > 0
where α, β , γ and δ are constants, then such boundary conditions are said to be homogeneous.
Otherwise, the boundary conditions are called nonhomogeneous.
149
4.2 The Heat Equation
Consider the following heat equation with nonhomogeneous boundary conditions.
Example 4.17. Let u(x, t) denote the temperature of a thin metal bar of length 10 that is insulated along the sides. Let the temperature along the length of the bar at time t = 0 be given by
u(x, 0) = x2
where 0 6 x 6 10.
Let the temperature of the end of the bar corresponding to x = 0 be held at 0 degrees; let the other
end be held at 100 degrees – that is
u(0, t) = 0
u(10, t) = 100
where t > 0
Determine the temperature u(x, t) of the bar by solving the following heat equation
∂u
∂ 2u
= c2 2
∂t
∂x
(4.60)
where c is a fixed constant that depends on the physical properties of the bar.
Answer: We will use a substitution of the form
u(x, t) = us(x, t) + U (x, t)
to convert this heat equation with nonhomogeneous boundary conditions to a heat equation with
homogeneous boundary conditions as in Example 4.14. We do this because the seperation of variables method fails in the case of nonhomogeneous boundary conditions.
Consider the following function
us(x, t) = 10x.
(4.61)
This function us can be interpreted as a linear increase in temperature from the end x = 0 which is
at temperature 0 to the end x = 10 at temperature 100.
We define our substitution U(x, t) by the equation
U (x, t) = u(x, t) − us(x, t),
(4.62)
the idea being to use us to ‘subtract’ the boundary condition from u(x, t) so that U (x, t) has zero
temperature at both ends, that is, the boundary conditions for U are homogeneous:
U (0, t) = u(0, t) − us(0, t)
=0−0
U (10, t) = u(10, t) − us(10, t)
= 100 − 10(10)
where t > 0
150
Partial Differential Equations
The function us was also chosen to be a solution of the heat equation. By differentiating (4.61) partially it is easy to see that
∂us
∂ 2u
= c2 2s
(4.63)
∂t
∂x
and because (4.63) holds, we have that U (x, t) also satisfies the heat equation: substituting
u(x, t) = U (x, t) + us(x, t)
into
∂ 2u
∂u
= c2 2
∂x
∂t
gives
∂(U + us)
∂ 2(U + us)
= c2
∂t
∂x2
⇒
∂U ∂us
∂ 2u
∂ 2U
+
= c2 2s + c2 2
∂t
∂t
∂x
∂x
(4.64)
and subtracting (4.63) from (4.64) gives
∂U
∂ 2U
= c2 2 .
∂t
∂x
(4.65)
(If the function us were not a solution of the heat equation, then the partial differential equation in
U obtained would not take the standard form of the heat equation (4.65) ).
From equation (4.62) we can also determine the initial condition for U(x, t)
u(x, 0) = 10x + U (x, 0)
U (x, 0) = x2 − 10x
⇒
for 0 6 x 6 10.
Therefore we have that the substitution (4.62) converts the heat equation
∂ 2u
∂u
= c2 2
∂x
∂t
subject to the nonhomogeneous boundary conditions
u(0, t) = 0
u(10, t) = 100
where t > 0
and initial condition
u(x, 0) = x2
where 0 6 x 6 10
to another heat equation
∂ 2U
∂U
= c2 2
∂x
∂t
subject to the homogeneous boundary conditions
U (0, t) = 0
U (10, t) = 0
where t > 0
and initial condition
U (x, 0) = x2 − 10x
where 0 6 x 6 10
(4.66)
151
4.2 The Heat Equation
Now the heat equation (4.66) has boundary conditions exactly of the type considered in Example
4.14 and so the technique of solving (4.66) is exactly as the technique used to solve Example 4.14.
We use the method of seperation of variables and therefore assume that the solution U (x, t) of (4.66)
takes the form
U (x, t) = X(x)T (t).
Substituting this into heat equation (4.66) gives
∂
∂2
(X(x)T (t)) = c2 2 (X(x)T (t))
∂t
∂x
X ′′(x)
T ′(t)
=
=k
⇒ 2
X(x)
c T (t)
and, exactly as in Example 4.14, the boundary conditions for U (x, t) imply that the only non-zero
solutions are obtained when k = − a2
T ′(t)
X ′′(x)
=
= − a2
2
c T (t)
X(x)
from which we obtain the two ordinary differential equations
T ′(t) + c2a2T (t) = 0
(4.67)
X ′′(x) + a2X(x) = 0
(4.68)
By using the D-operator method, the general solution of (4.67) is
2 2
T (t) = Ce−c a t
and the general solution of (4.68) is
X(x) = D cos(ax) + E sin(ax).
And so the solutions of the heat equation take the form
2 2
U (x, t) = e−c a t(A cos (ax) + B sin (ax))
(4.69)
where the arbitrary constants A = CD and B = CE.
We impose the boundary conditions on equation (4.69). From the first boundary condition of (4.66)
we get
2 2
U(0, t) = e−c a t(A cos(0) + B sin (0)) ≡ 0
2 2
⇒ Ae−c a t ≡ 0
⇒ A = 0.
Substituting A = 0 into (4.69) implies that our solutions are of the form
2 2
U (x, t) = Be−c a tsin(ax).
(4.70)
Imposing the second boundary condition on equation (4.70) gives
2 2
U (10, t) = Be−c a tsin(10a) ≡ 0
⇒ sin(10a) = 0
and so the second boundary condition forces
10a = kπ
where k = 1, 2, 3, Substituting each of these different values of a into equation (4.70) implies that
e
−c2
π 2
t
10
sin(
2π
3π
2πx
3πx
πx
−c2( 10 )2t
−c2( 10 )2t
sin(
sin(
), e
), e
), 10
10
10
are each solutions of the heat equation . So from by the Superposition Principle we have that
U(x, t) =
∞
X
n=1
bne
−c2
nπ 2
t
10
nπx sin
10
(4.71)
152
Partial Differential Equations
is a solution of the heat equation (4.66) which, by construction, satisfies the boundary conditions.
We use the initial condition
U (x, 0) = x2 − 10x
where 0 6 x 6 10
to determine the constants bn of equation (4.71). Setting t = 0 in (4.71) gives
U(x, 0) =
∞
X
bne
−c2
n=1
⇒
∞
X
n=1
nπ 2
0
10
nπx sin
= x2 − 10x when 0 6 x 6 10
10
nπx bnsin
= x2 − 10x when 0 6 x 6 10
10
(4.72)
The left side of equation (4.72) is a Fourier sine series that represents the function x2 − 10x on the
interval 0 6 x 6 10. From Lemma 3.23 on page 128, the coefficients bn are given by
Z
nπx
2 L
bn =
f (x)sin
dx
L
L 0
Z 10
nπx 2
(x2 − 10x)sin
dx
⇒ bn =
10
10 0
Integrating by parts with
nπx dx
du = sin
10 10
nπx dv = (2x − 10) dx
u=−
cos
nπ
10
Z
nπx 10
nπx 10
1
2
10(x2 − 10x)
bn =
cos
+
(2x − 10) cos
dx
−
10
5
nπ 0
10
nπ
0
Z 10
2
nπx
=0+
(2x − 10) cos
dx
nπ 0
10
v = x2 − 10x
gives
Integrating by parts again with
nπx dx
du = cos
10
10 nπx dv = 2dx
u=
sin
10
nπ
!
Z
10
10
20 nπx 10(2x − 10) nπx 2
−
sin
sin
dx
bn =
nπ
10
10
nπ
nπ
0
0
nπx 10
2 200
=
cos
nπ n2π 2
10
0
200
2 200( − 1)n
− 2 2
=
nπ
n 2π 2
nπ


0
when n is even

=
800

 − 3 3 when n is odd
nπ
v = 2x − 10
gives
Substituting these values for bn into (4.71) and writing the odd values of n as 2k + 1 gives our solution for U (x, t)
∞
(2k+1)π 2
X
−c2
t
(2k + 1)πx
800
10
e
sin
.
U (x, t) = −
10
(2k + 1)3π 3
k=0
and since
u(x, t) = us(x, t) + U (x, t)
we obtain the solution of our nonhomogeneous problem
153
4.2 The Heat Equation
u(x, t) = 10x −
∞
X
k=0
−c2
800
e
(2k + 1)3π 3
(2k+1)π
10
2
t
(2k + 1)πx
sin
10
(4.73)
In Example 4.17, holding one end at temperature zero in effect creates a heat sink. Holding the
other end at temperature 100 creates a heat source, and we should expect that as time t → ∞ that
the presence of this source and sink to dominate the temperature distribution along the length of the
bar. As this source and sink does not change with time (because the temperature at the ends of the
bar are fixed) we should also expect that the temperature distribution along the bar to stabilize,
that is, each point in the bar should as time t → ∞ to have a temperature that is independent of
time. This physical expectation agrees with the solution (4.73) of Example 4.17 as
!
∞
(2k+1)π 2
X
t
−c2
(2k + 1)πx
800
10
sin
e
lim u(x, t) = lim 10x −
10
(2k + 1)3π 3
t→∞
t→∞
k=0
2
!
∞
(2k+1)π
X
−c2
t
800
(2k + 1)πx
10
= lim 10x − lim
e
sin
(2k + 1)3π 3
10
t→∞
t→∞
k=0
2
πx
800 −c2 10π 2t
2πx
800 −c2( 2π
)
t
10
sin( ) − lim 3 3 e
= lim 10x − lim 3 e
sin(
)−
10
π
10
3
π
t→∞
t→∞
t→∞
= lim 10x − 0 − 0 − t→∞
= 10x
and note
lim u(x, t) = 10x
t→∞
can be interpreted as meaning that the temperature at any point x on the bar (where 0 6 x 6 10)
will eventually reach the value of 10x. Also note that the temperature distribution
10x
is independent of time.
154
Partial Differential Equations
4.3 The Wave Equation
4.3.1 A derivation of the wave equation
Consider an elastic string that is stretched taut along the x-axis, with one end of the string fixed at
the point x = 0 and the other end fixed at the point x = L.
The string is then distorted at time t = 0
and because the string is under tension, when the string is released it will oscillate about the x −
axis with the endpoints of the string being stationary.
We make the assumption that each point on the string travels vertically under this oscillation.
Therefore we can identify each point on the string by an x − value and we let the displacement of
this point from the horizontal axis at time t be given by the function value u(x, t).
Note that as the ends of the string (which correspond to the points x = 0 and x = L) are fixed we
have the boundary conditions
u(0, t) = 0
for all t > 0.
(4.74)
u(L, t) = 0
If the distortion of the string at time t = 0 is given by the function f (x) where 0 6 x 6 L, then we
have an initial condition
u(x, 0) = f (x) where 0 6 x 6 L.
(4.75)
155
4.3 The Wave Equation
Now if
u(x, t) = displacement of point x at time t
then
∂u
(x, t) = velocity of point x at time t.
∂t
Therefore if the velocity of each point x on the string is given by the function g(x) where 0 6 x 6 L,
then we have a second initial condition
∂u
(x, 0) = g(x) where 0 6 x 6 L.
∂t
(4.76)
Summarizing, our physical problem is as follows. We have a string that is stretched
horizontally between two points that are at a distance L apart. The string has an initial distortion given by equation (4.75) and an initial velocity given by (4.76). Each
point on the string can be identified with a point x where 0 6 x 6 L. We wish to find
a function u(x, t) that measures the displacement of each point on the string from the
horizontal axis at time t. This function u(x, t) will be a solution of the wave equation.
To derive the wave equation we consider, at time t, a small segment AB of the string which consists of the points that correspond to the x − values that lie between x and x + ∆x where ∆x represents a small positive number:
In the above diagram T1 denotes the tension in the string at the point A; T2 denotes the tension at
point B. When ∆x is small enough we may assume that the segment of string A B does not move
horizontally, this implies that the net horizontal force on the segment A B is zero. By considering
the horizontal components of the tensions T1 and T2 we have
T1cos α = T2 cosβ
and so T1cos α and T2 cosβ have a common value which we call T , therefore
T1cos α = T2 cosβ = T .
(4.77)
For ∆x small enough we may also assume that the segment of string A B moves as a particle. The
displacement of this particle from the horizontal axis is given by u(x, t), therefore the velocity of this
∂u
∂ 2u
particle in the vertical direction is ∂t (x, t) and the acceleration in the vertical direction is ∂t2 (x, t).
By considering the vertical components of the tensions T1 and T2, from Newton’s Second Law we
have
∂ 2u
T2 sin β − T1sin α = (ρ∆x) 2
(4.78)
∂t
156
Partial Differential Equations
where ρ is the mass per unit length of the string. Dividing equation (4.78) by the value T from
equation (4.77) we have
T2 sin β T1sin α
ρ∆x ∂ 2u
−
=(
)
T
T
T ∂t2
and using the expressions for T from equation (4.77) we have
T2 sin β T1sin α
ρ∆x ∂ 2u
−
=(
)
T2cosβ T1cos α
T ∂t2
ρ∆x ∂ 2u
⇒ tan β − tan α = (
)
.
T ∂t2
(4.79)
In the following diagram the line segment B C is tangent to the curve u(x, t) at the point B (note
that we regard the t − variable as being fixed).
By the definition of slope we have
slope of tangent BC =
CD
= tan β.
BD
However, from the definition of the partial derivative
slope of tangent BC =
∂u
(x + ∆x, t)
∂x
and therefore the slope of the tangent at B is
∂u
(x + ∆x, t) = tan β.
∂x
Similarly the slope of the tangent at A is
∂u
(x, t) = tan α
∂x
and from the equations of (4.79) we have
∂u
∂u
ρ∆x ∂ 2u
(x + ∆x, t) −
(x, t) = (
)
.
∂x
∂x
T ∂t2
Rearranging this equation gives
∂u
∂u
(x + ∆x, t) − ∂x (x, t)
∂x
∆x
ρ ∂ 2u
=( ) 2
T ∂t
(4.80)
and we take the limit of (4.80) as ∆x → 0 to get
lim
∆x→0
∂u
∂u
(x + ∆x, t) − ∂x (x, t)
∂x
∆x
!
ρ ∂ 2u
=( ) 2.
T ∂t
(4.81)
157
4.3 The Wave Equation
By definition of the partial derivative
lim
∂u
∂u
(x + ∆x, t) − ∂x (x, t)
∂x
∆x
∆x→0
!
=
∂
∂x
∂ 2u(x, t)
∂u
(x, t) =
∂x2
∂x
and so from equation (4.81) we obtain
∂ 2u(x, t)
ρ ∂ 2u
=( ) 2
2
∂x
T ∂t
(4.82)
and as ρ and T are positive constants we may write equation (4.82) in the standard form of the
wave equation
∂ 2u
∂ 2u
= c2 2
2
∂x
∂t
where c is a constant.
4.3.2 Solution of the wave equation by seperation of variables
The solution of the wave equation by the method of seperation of variables is similar to the solution
of the heat equation by seperation of variables that was done in Section 4.2.2. In solving the wave
equation we shall
−
assume a solution of the form
−
substitute this solution into the wave equation to obtain
u(x, t) = X(x)T (t)
T ′′(t) X ′′(x)
=
=k
X(x)
c2T (t)
−
consider the cases of k = a2, 0, − a2 and use boundary conditions to determine which of these
cases of k lead to nonzero solutions of the wave equation
−
use the Superposition Principle to write these non-zero solutions in expressions that
resemble4.5
∞ X
nπct
nπx nπct
+ bnsin
sin
u(x, t) =
ancos
L
L
L
n=1
−
use initial conditions and the half-range cosine/sine formulae
Z
nπx
2 L
dx
f (x)sin
an =
L
L 0
Z
nπcbn 2 L
nπx
g(x)sin
=
dx
L
L 0
L
to determine the constants an and bn, where f (x) and g(x) describe the initial displacement
and velocity.
4.5. solutions may not always take this form
158
Partial Differential Equations
We illustrate the solution of the wave equation by the following example.
Example 4.18. Solve the wave equation
∂ 2u ∂ 2u
= 2
∂t2
∂x
subject to the boundary conditions
(4.83)
u(0, t) = 0
u(π, t) = 0
and initial conditions
1
u(x, 0) = sin x − sin 2x
2
∂u
(x, 0) = 0
∂t
where 0 6 x 6 π.
Answer: We use the method of seperation of variables and therefore assume that our solution u(x, t)
takes the form
u(x, t) = X(x)T (t).
(4.84)
Substituting the solution (4.84) into wave equation (4.83) gives
∂2
∂2
(X(x)T (t)) = 2 (X(x)T (t))
2
∂t
∂x
⇒ X(x)T ′′(t) = X ′′(x)T (t)
⇒
T ′′(t) X ′′(x)
=
X(x)
T (t)
T ′′(t)
(4.85)
X ′′(x)
and the only way for equation (4.85) to hold is if T (t) and X(x) are both constant functions and
are equal to the same constant k. So we have
T ′′(t) X ′′(x)
=
=k
X(x)
T (t)
(4.86)
where the constant k can be zero, negative or positive, that is
k = 0, − a2 or a2
where a 0. We now use the boundary conditions
u(0, t) = 0
u(L, t) = 0
where t > 0
(4.87)
to show that the only non-zero solutions of the wave equation (4.83) with the boundary conditions
(4.87) occur when k = − a2.
k=0
If k = 0 then from equation (4.84) we obtain two ordinary differential equations
T ′′(t) = 0
X ′′(x) = 0.
Solving these gives
T (t) = A ′t + B ′
X(x) = C ′x + D ′
where A ′, B ′,C ′ and D ′ are constants.
(4.88)
159
4.3 The Wave Equation
Substituting equations (4.88) into equation (4.84) gives
u(x, t) = (C ′x + D ′)(A ′t + B ′)
(4.89)
From the first boundary condition of (4.87) we get
u(0, t) = D ′(A ′t + B ′) ≡ 0
for all values of t > 0, which implies that
D ′ = 0.
The second boundary condition gives
u(π, t) = πC ′(A ′t + B ′) ≡ 0
for all values of t > 0, which implies that
C ′ = 0.
As the boundary conditions imply that
C′ = 0
D′ = 0
we have that when k = 0, the corresponding solution (4.89) is identically equal to zero.
k = a2
If k = a2 then from equation (4.84) we obtain two ordinary differential equations
T ′′(t) − a2T (t) = 0
X ′′(x) − a2X(x) = 0.
Solving these gives
T (t) = A ′′cosh(at) + B ′′sinh(at)
X(x) = C ′′cosh(ax) + D ′′sinh(ax)
(4.90)
where A ′′, B ′′,C ′′ and D ′′ are constants.
Substituting equations (4.90) into equation (4.84) gives
u(x, t) = (C ′′cosh(ax) + D ′′sinh(ax))(A ′′cosh(at) + B ′′sinh(at))
From the first boundary condition of (4.87) we get
u(0, t) = (C ′′(1) + D ′′(0))(A ′′cosh(at) + B ′′sinh(at)) ≡ 0
for all values of t > 0, which implies that
C ′′ = 0.
The second boundary condition gives
u(π, t) = (D ′′sinh(aπ))(A ′′cosh(at) + B ′′sinh(at)) ≡ 0
for all values of t > 0, and as sinh(aπ) 0 when a 0, this implies that
D ′′ = 0.
As the boundary conditions imply that
C ′′ = 0
D ′′ = 0
we have that when k = a2, the corresponding solution (4.91) is identically equal to zero.
(4.91)
160
Partial Differential Equations
k = − a2
The last possibility of k = − a2 will lead to non-zero solutions. Subsituting k = − a2 into equation
(4.86) gives
T ′(t) X ′′(x)
=
= − a2
T (t)
X(x)
from which we obtain the two ordinary differential equations
T ′′(t) + a2T (t) = 0
(4.92)
X ′′(x) + a2X(x) = 0
(4.93)
By using the D-operator method, the general solution of (4.92) is
T (t) = A cos(at) + B sin(at)
and the general solution of (4.93) is
X(x) = C cos(ax) + D sin(ax).
So we have solutions of the wave equation of the form
u(x, t) = (A cos(at) + B sin(at))(C cos (ax) + D sin (ax)).
(4.94)
From the first boundary condition we get
u(0, t) = (A cos(at) + B sin(at))(C(1) + D(0)) ≡ 0
⇒ C(A cos(at) + B sin(at)) ≡ 0
⇒ C =0
and therefore the solutions take the form
u(x, t) = (a ′ cos(at) + b ′ sin(at))sin (ax)
(4.95)
where a ′ = AD and b ′ = BD.
Imposing the second boundary condition on equation (4.95) gives
u(π, t) = (a ′ cos(at) + b ′ sin(at))sin (aπ) ≡ 0
⇒ sin(aπ) = 0
and so the second boundary condition forces
a=k
where k = 1, 2, 3, (note we do not consider k = 0 as this leads to the zero solution; k = − 1, − 2, do not give solutions
that are linearly independent of those corresponding to k = 1, 2, ). Substituting each of these different values of a into equation (4.95) implies that
(a1cos(t) + b1sin(t))sin(x) , (a2cos(2t) + b2sin(2t))sin(2x) , (a3cos(3t) + b3sin(3t))sin(3x), are each solutions of the wave equation (4.83). So from by the Superposition Principle II on page
133 we have that
∞
X
(ancos(nt) + bnsin(nt))sin(nx)
(4.96)
u(x, t) =
n=1
is a solution of the wave equation (4.83) which, by construction, satisfies the boundary conditions.
Finally, we use the given initial conditions
1
u(x, 0) = sin x − sin 2x
2
∂u
(x, 0) = 0
∂x
where 0 6 x 6 π.
161
4.3 The Wave Equation
to determine the coefficients an , bn of the solution (4.96). Setting t = 0 in (4.96) gives
u(x, 0) =
∞
X
n=1
⇒
1
(ancos(0) + bnsin(0))sin(n x) = sin x − sin 2x
2
∞
X
n=1
1
ansin(n x) = sin x − sin 2x
2
when 0 6 x 6 π
when 0 6 x 6 π
(4.97)
1
The left side of equation (4.97) is a Fourier sine series that represents the function sin x − 2 sin 2x on
the interval 0 6 x 6 π. Such a sine series is called the sine half-range expansion of the function
1
sin x − 2 sin 2x on the interval 0 6 x 6 π and from Lemma 3.23 on page 128, the coefficients an of
the left side of (4.97) are given by
Z
2 L
nπx
an =
f (x)sin
dx where n = 1, 2, L 0
L
Z 2 π
1
⇒ an =
sin x − sin 2x sin (nx) dx where n = 1, 2, π 0
2
Now if n is an integer, then
Z
π
sin(nx)sin (nx) dx =
0
=
π
Z
Z0
0
=
and when n m are both integers
Z π
Z
sin(nx)sin (mx) dx =
0
0
π
π
2
π
sin2(nx) dx
1 − cos(2nx)
dx
2
cos(m − n)x − cos(m + n)x
dx
2
= 0.
Therefore we have
Z 1
2 π
sin x − sin 2x sin x dx
a1 =
2
π 0
Z π
Z
1 π
2
sin x sin x dx −
sin 2x sin x dx
=
π 0
π 0
2 π 1
= . − .0
π 2 π
=1
and similarly
Z 1
2 π
sin x − sin 2x sin 2x dx
a2 =
2
π 0
Z π
Z
2
1 π
=
sin x sin2x dx −
sin 2x sin2x dx
π 0
π 0
2
1 π
= .0 − .
π
π 2
1
=− .
2
For ak where k > 2 we have
Z 2 π
1
sin x − sin 2x sin kx dx
π 0
2
Z
Z
2 π
1 π
=
sin x sin kx dx −
sin 2x sin kx dx
π 0
π 0
2
1
= .0 − .0
π
π
=0
ak =
162
Partial Differential Equations
and therefore we have determined the an
a1 = 1 a2 = −
1
2
and an = 0 for all n > 2.
(4.98)
Differentiating the solution (4.96) partially with respect to t gives
∞
X
∂u
(x, t) =
( − nansin(nt) + nbncos(nt))sin(nx)
∂t
n=1
and from the second initial condition we have
∞
which implies
X
∂u
( − nansin(0) + nbncos(0))sin(nx) ≡ 0
(x, 0) =
∂t
when 0 6 x 6 π
n=1
∞
X
n=1
nbnsin(nx) ≡ 0
(4.99)
The left side of equation (4.99) is a Fourier sine series that represents the function 0 on the interval
0 6 x 6 π and from Lemma 3.23 on page 128, the coefficients nbn of the left side of (4.99) are given
by
Z
2 L
nπx
dx where n = 1, 2, nbn =
f (x)sin
L 0
L
Z
2 π
⇒ nbn =
(0)sin (nx) dx where n = 1, 2, π 0
⇒ nbn = 0
⇒ bn = 0
where n = 1, 2, (4.100)
Substituting the values of the coefficients an and bn given by (4.98) and (4.100) into (4.96) we obtain
our solution
1
u(x, t) = cos(t)sin(x) − cos(2t)sin(2x)
2
of the wave equation (4.83) that satisfies the given boundary and initial conditions.
163
4.4 Laplace’s equation
4.4 Laplace’s equation
Laplace’s equation is a partial differential equation that does not involve the time variable t. In
Cartesian coordinates, the form of Laplace’s equation depends on dimension:
Laplace ′s equation in 1 dimension
d 2u
=0
dx2
Laplace ′s equation in 2 dimensions
∂ 2u ∂ 2u
+
=0
∂x2 ∂y 2
Laplace ′s equation in 3 dimensions
∂ 2u ∂ 2u ∂ 2u
+
+
=0
∂x2 ∂y 2 ∂y 2
Laplace’s equation arises when one considers solutions of physical problems that are independent of
time. Solutions that are independent of time are called steady state solutions.
Example 4.19. In Example 4.17 we showed on page 150 that
us(x, t) = 10x
is a solution of the heat equation
∂ 2u
∂u
= c2 2
∂x
∂t
subject to the nonhomogeneous boundary conditions
u(0, t) = 0
u(10, t) = 100
where t > 0
Notice that the solution us does not involve the t variable, so us is an example of a steady-state solution. As us is a solution of this heat equation we clearly have
∂ 2u
∂us
= c2 2s .
∂x
∂t
(4.101)
Furthermore as us is independent of the time variable t, when we differentiate partially with respect
to t we have
∂us
=0
(4.102)
∂t
and substituting (4.102) into (4.101) we have that, after dividing by c2, that
∂ 2us
=0
∂x2
and as us is a function of x only we have that
d2us
=0
dx2
which is Laplace’s equation in one dimension. So we see that us is an example of a steady-state solution of a one-dimensional heat equation being a solution of Laplace’s equation in one-dimension.
To be precise, us is a solution of the Laplace’s equation together with specified boundary conditions:
164
Partial Differential Equations
d2u
=0
dx2
subject to the boundary conditions
u(0) = 0
u(10) = 100
Finally, recall that the limit of the solution of the heat equation in Example 4.17
lim u(x, t) = lim
t→∞
t→∞
10x −
= 10x
= us(x, t)
∞
X
k=0
−c2
800
e
3
3
(2k + 1) π
(2k+1)π
10
2
t
!
(2k + 1)πx
sin
10
So we see that the steady-state solution us of the the heat equation in Example 4.17 gives the temperature distribution that the metal bar eventually stabilizes to. As us is a solution of a Laplace’s
equation, we have that the limiting temperature distribution of the solution of a heat equation can
sometimes be obtained as a solution of a corresponding Laplace’s equation with the appropriate
boundary conditions.
We see an example of this in the following
Example 4.20. From Example 4.14 we saw the solution of the following heat equation problem
Let u(x, t) denote the temperature of a thin metal bar of length 1 that is insulated
along the sides. Let the temperature along the length of the bar at time t = 0 be
given by
u(x, 0) = 4x(1 − x)
where 0 6 x 6 1.
The ends of the bar are kept at a constant temperature of 0, that is
u(0, t) = 0
u(1, t) = 0
is given by
u(x, t) =
∞
X
k=0
where t > 0.
2
2
32
e−c ((2k+1)π) tsin((2k + 1)πx)
3
3
(2k + 1) π
and this temperature function eventually stablilizes to
lim u(x, t) = 0
t→∞
(4.103)
as we would expect because all the initial heat in the bar flows out of the ends of the bar which are
at zero temperature. Notice that this limiting temperature distribution of zero (4.103) is identical to
the solution of a corresponding one-dimensional Laplace’s equation with the same boundary conditions as the heat equation
d2u
=0
dx2
subject to the boundary conditions
u(0) = 0
u(1) = 0
165
4.4 Laplace’s equation
because the solution of the above Laplace’s equation is easily shown to be zero:
and the boundary conditions
imply that
d2u
=0
dx2
⇒ u(x) = Ax + B
u(0) = 0
u(1) = 0
A=B =0
hence
u = 0.
which is the limiting temperature distribution of (4.103).
The previous two examples provide evidence that Laplace’s equation in one dimension can be interpreted as a steady-state heat flow problem in one dimension, that is, a problem of finding a
steady-state solution of a one dimensional heat equation that satisfies some given boundary conditions.
In this section we shall be interested in solving the two dimensional Laplace’s equation over a rectangular domain. The following lemma implies that Laplace’s equation in two dimensions can also be
interpreted as a steady-state heat flow problem in two dimensions, that is, a problem of finding a
steady-state solution of a two dimensional heat equation that satisfies certain conditions on the
boundary of the domain.
Lemma 4.21. Let us be a steady-state solution of the two dimensional heat equation
2
∂ u ∂ 2u
∂u
.
= c2
+
∂x2 ∂y 2
∂t
(4.104)
Then us is a solution of Laplace’s equation in two dimensions
∂ 2u ∂ 2u
+
= 0.
∂x2 ∂y 2
Proof. As us is a solution of this heat equation we clearly have
2
∂ us ∂ 2us
∂us
= c2
+
∂x2
∂y 2
∂t
(4.105)
As us is steady-state, it is independent of the time variable t and therefore
∂us
=0
∂t
(4.106)
and by substituting (4.106) into (4.105) and dividing by c2 we have that
∂ 2us ∂ 2us
+
=0
∂y 2
∂x2
and therefore us is a solution of Laplace’s equation in two dimensions.
166
Partial Differential Equations
4.4.1 Solving Laplace’s equation by seperation of variables
We illustrate the solution of Laplace’s equation over a rectangular domain with prescribed boundary
conditions by the following example.
Example 4.22. By solving Laplace’s equation
∂ 2u ∂ 2u
+
=0
∂x2 ∂y 2
(4.107)
in the square domain
{(x, y)| 0 6 x 6 1 and 0 6 y 6 1},
find the steady-state temperature u(x, y) of a thin square metal plate of length 1 which has each of
its four sides at temperatures that are given by the following boundary conditions
u(0, y) = 0
u(1, y) = 0
where 0 6 x 6 1 and 0 6 y 6 1.
u(x, 0) = 0
u(x, 1) = x(1 − x)
(4.108)
Answer: The following diagram illustrates the boundary conditions around the square plate:
We solve this Laplace’s equation by using seperation of variables and therefore assume that our solution u(x, y) takes the form
u(x, y) = X(x)Y (y).
(4.109)
Substituting the solution (4.109) into Laplace’s equation (4.107) gives
∂2
∂2
(X(x)Y (y)) + 2 (X(x)Y (y)) = 0
2
∂x
∂y
⇒ X ′′(x)Y (y) + X(x)Y ′′(y) = 0
⇒
Y ′′(y)
X ′′(x)
=−
Y (y)
X(x)
(4.110)
167
4.4 Laplace’s equation
and the only way for equation (4.85) to hold is if
X ′′(x)
and
X(x)
and are equal to the same constant k. So we have
−
Y ′′(y)
are both constant functions
Y (y)
Y ′′(y)
X ′′(x)
=−
=k
Y (y)
X(x)
(4.111)
where the constant k can be zero, negative or positive, that is
k = 0, − a2 or a2
where a 0. As with the heat and wave equations, we use the boundary conditions (4.108) to determine which of the k = 0, − a2 or a2 will give non-zero solutions.
k=0
If k = 0 then from equation (4.111) we obtain two ordinary differential equations
X ′′(x) = 0
Y ′′(y) = 0.
Solving these gives
X(x) = A ′x + B ′
Y (y) = C ′ y + D ′
(4.112)
where A ′, B ′,C ′ and D ′ are constants.
Substituting equations (4.112) into equation (4.109) gives
u(x, y) = (A ′x + B ′)(C ′ y + D ′)
and it is easy to check that the boundary conditions
u(0, y) = 0
u(1, y) = 0
imply that A ′ = B ′ = 0 and hence the only solution u(x, y) corresponding to k = 0 is the zero solution
u(x, y) = 0.
k = a2
If k = a2 then from equation (4.111) we obtain two ordinary differential equations
X ′′(x) − a2X(x) = 0
Y ′′(y) − a2Y (y) = 0.
Solving these gives
X(x) = A ′′cosh(ax) + B ′′sinh(ax)
Y (y) = C ′′cosh(ay) + D ′′sinh(ay)
where A ′′, B ′′,C ′′ and D ′′ are constants.
Substituting equations (4.113) into equation (4.111) gives
u(x, y) = (A ′′cosh(ax) + B ′′sinh(ax))(C ′′cosh(ay) + D ′′sinh(ay))
and again it is easy to check that the boundary conditions
u(0, y) = 0
u(1, y) = 0
(4.113)
168
Partial Differential Equations
imply that A ′′ = B ′′ = 0 and hence the only solution u(x, y) corresponding to k = a2 is the zero solution
u(x, y) = 0.
k = − a2
The last possibility of k = − a2 will lead to non-zero solutions. Subsituting k = − a2 into equation
(4.111) gives
X ′′(x)
Y ′′(y)
=−
= − a2
X(x)
Y (y)
from which we obtain the two ordinary differential equations
X ′′(x) + a2X(x) = 0
Y ′′(y) − a2Y (y) = 0
Solving these gives
X(x) = A cos(ax) + B sin(ax)
Y (y) = C cosh(ay) + D sinh (ay)
(4.114)
where A, B, C and D are constant. Substituting (4.114) into (4.109) gives
u(x, y) = (A cos(ax) + B sin(ax))(C cosh(ay) + D sinh (ay)).
From the first boundary condition
u(0, y) = 0
of (4.108) we have
u(0, y) = (A cos(0) + B sin(0))(C cosh(ay) + D sinh (ay)) ≡ 0
⇒ A(C cosh(ay) + D sinh (ay)) ≡ 0
⇒
A=0
and therefore our solution u(x, y) takes the form
u(x, y) = sin(ax)(a ′cosh(ay) + b ′sinh(ay))
(4.115)
where a ′ = BC and b ′ = BD.
Imposing the second boundary condition
u(1, y) = 0
on the solution (4.115) gives
u(1, y) = sin(a)(a ′cosh(ay) + b ′sinh(ay)) ≡ 0
⇒
and so the second boundary condition forces
a = kπ
sin(a) = 0
where k = 1, 2, 3, Substituting each of these different values of a into (4.115) implies that
sin(πx)(a1cosh(πy) + b1sinh(πy))
sin(2πx)(a2cosh(2πy) + b2sinh(2πy))
sin(3πx)(a3cosh(3πy) + b3sinh(3πy))
are each solutions of Laplace’s equation. So from the Superposition Principle we have that
u(x, y) =
∞
X
sin(nπx)(ancosh(nπy) + bnsinh(nπy))
(4.116)
n=1
is a general form of the solution of Laplace’s equation which, by construction, satisfies the first two
boundary conditions of (4.108).
169
4.4 Laplace’s equation
Finally, we use the other two boundary conditions
u(x, 0) = 0
u(x, 1) = x(1 − x)
when 0 6 x 6 1
to determine the coefficients an , bn of the solution (4.116). From
u(x, 0) = 0
we have
u(x, 0) =
∞
X
sin(nπx)(ancosh(0) + bnsinh(0)) = 0
n=1
⇒
⇒
∞
X
sin(nπx)(an(1) + bn(0)) = 0
n=1
∞
X
ansin(nπx) = 0
when 0 6 x 6 1.
(4.117)
n=1
The left side of equation (4.117) is a Fourier sine series that represents the function 0 on the interval
0 6 x 6 1 and from Lemma 3.23 on page 128, the coefficients an on the left side of (4.117) are given
by
Z
2 L
nπx
an =
f (x)sin
dx where n = 1, 2, L 0
L
Z
2 1
⇒ an =
(0)sin nπx dx where n = 1, 2, 1 0
⇒ an = 0 where n = 1, 2, and so we have that our solution (4.116) of Laplace’s equation simplifies to
u(x, y) =
∞
X
bnsin(nπx)sinh(nπy).
(4.118)
n=1
From the fourth boundary condition
u(x, 1) = x(1 − x)
we have
u(x, 1) =
∞
X
n=1
⇒
∞
X
n=1
bnsin(nπx)sinh(nπ ) = x(1 − x) when 0 6 x 6 1
(bnsinh (nπ))sin(nπx) = x(1 − x) when 0 6 x 6 1.
(4.119)
The left side of equation (4.119) is a Fourier sine series with coefficients
bnsinh(nπ)
that represents the function x(1 − x) on the interval 0 6 x 6 1 and from Lemma 3.23 on page 128,
the coefficients bnsinh(nπ) on the left side of (4.119) are given by
Z
nπx
2 1
dx where n = 1, 2, x(1 − x)sin
bnsinh(nπ) =
1
1 0
Integrating
Z
1
2
by parts with
0
v = x(1 − x)
gives
x(1 − x)sin (nπx)dx
du = sin(nπx)dx
cos (nπx)
dv = (1 − 2x) dx
u=−
nπ
1
Z 1
(1 − 2x)cos (nπx)
x(1 − x)cos (nπx)
dx
bnsinh(nπ) = 2 −
+2
nπ
nπ
0
0
Z 1
(1 − 2x)cos (nπx)
=0+2
dx
nπ
0
170
Partial Differential Equations
Integrating by parts again with
cos (nπx)
dx
nπ
sin (nπx)
dv = − 2 dx
u=
n 2π 2
1
Z 1
(1 − 2x)sin (nπx)
2sin (nπx)
bnsinh(nπ) = 2
+
2
dx
n 2π 2
n 2π 2
0
0
1
2cos (nπx)
=0+2 −
n 3π 3
0
2( − 1)n
2
=2 −
+ 3 3
n3π 3
nπ


when n is even
 0
=
8

 3 3 when n is odd
nπ

0
when n is even


bn =
8

when n is odd
 3 3
n π sinh(nπ)
v = (1 − 2x)
gives
and therefore
du =
Substituting these values for bn into (4.118) and writing the odd values of n as 2k + 1 gives our solution for u(x, y)
u(x, y) =
∞
X
k=0
8
sin((2k + 1)πx)sinh((2k + 1)πy)
(2k + 1)3π 3sinh((2k + 1)π)
(4.120)
which gives the steady-state temperature distribution of the square plate with the given boundary
conditions.
The following contour plot, drawn with the computer program Maple, shows a plot of the solution (4.120) that represents a temperature distribution of the square plate (for example, all points on
the curve labelled u = .04 have temperature equal to .04). Note that the maximum temperature of
.25 on the square plate is achieved at the point
(x, y) = (.5, 1)
which lies on the top edge of the plate.
Figure 4.1. Contour plot showing steady-state solution to Example 4.22
171
4.5 Laplace’s equation in polar coordinates
4.5 Laplace’s equation in polar coordinates
In the last section we used Laplace’s equation
∂ 2 u ∂ 2u
+
=0
∂x2 ∂y 2
(4.121)
to solve a steady-state heat flow problem in a square plate. If we wish to solve similar problems in
circular plates, it is sometimes easier to work with polar coordinates (r, θ) than with Cartesian
coordinates (x, y).
Recall that a point may be located in the plane by its Cartesian coordinates (x, y). The location
of this same point may also be given by the polar coordinates (r, θ)
Figure 4.2.
where r is the distance of the point (x, y) from the origin and θ is the angle that the line connecting
the point (x, y) with the origin makes with the x − axis. By considering the above right-angled triangle, clearly we have
x = r cos θ
(4.122)
y = r sin θ.
(4.123)
u(x, y)
(4.124)
Let
be a solution of the two-dimensional Laplace’s equation (4.121). Then if we replace x in the solution
u(x, y) by equation (4.122) and replace y by equation (4.123) we obtain a function
u(r cos θ, r sin θ)
(4.125)
in the variables r and θ. We will prove in Lemma 4.27 that the function (4.125) is a solution of the
partial differential equation
∂ 2u 1 ∂u 1 ∂ 2u
+
+
=0
(4.126)
∂r2 r ∂r r2 ∂θ 2
and the equation (4.126) is called Laplace’s equation in polar coordinates.
The opposite is also true (we will not prove this). If w(r, θ) is a solution of Laplace’s equation in
polar coordinates, then by using the substitutions
1
r = (x2 + y 2) 2
y
θ = tan−1
.
x
we can ‘convert’ w(r, θ) into a function of the variables x and y
y
1
)
w((x2 + y 2) 2 , tan−1
x
(4.127)
(4.128)
172
Partial Differential Equations
and this function will be a solution of Laplace’s equation (4.121) in Cartesian coordinates.
Example 4.23. By differentiating partially, it is easy to check that the function u(x, y) = x is a
solution of Laplace’s equation
∂ 2u ∂ 2 u
+
= 0.
∂x2 ∂y2
Expressing this solution u in polar coordinates gives
u(r, θ) = r cos θ
and this function of (r, θ) satisfies Laplace’s equation (4.121) in polar coordinates
∂ 2u 1 ∂u 1 ∂ 2u
+
+
∂r2 r ∂r r2 ∂θ2
2
∂
1 ∂
1 ∂2
= 2 (r cos θ) +
(r cos θ) + 2 2 (r cos θ)
∂r
r ∂r
r ∂θ
1
1
= 0 + (cos θ) + 2 ( − r cos θ)
r
r
=0
Recall the following basic result for derivatives of functions of one variable.
Lemma 4.24. (Chain Rule) If y = f (u) and u = g(x) are both differentiable functions then
y = f (g(x))
is a differentiable function of x and
dy dy du
=
·
dx du dx
Example 4.25. Let y = u10 and u = x2 + x. Then from Lemma 4.24
y = (x2 + x)10
is a differentiable function of x and
dy
= 10u9 · (2x + 1)
dx
= 10(x2 + x)9(2x + 1).
For functions of two variables, there is also a version of the Chain Rule which we will state in the
form that we shall need.
Lemma 4.26. Suppose that u(r, θ) is a differentiable function of r and θ. Let
r = g(x, y)
θ = h(x, y)
and suppose that the partial derivatives
exist. Then
∂r ∂r ∂θ
∂θ
, ,
and
∂x ∂y ∂x
∂y
u(g(x, y), h(x, y))
is a differentiable function of x and y and
∂u ∂u ∂r ∂u ∂θ
=
·
+
·
∂x ∂r ∂x ∂θ ∂x
∂u ∂u ∂r ∂u ∂θ
=
·
+
·
∂y ∂r ∂y ∂θ ∂y
173
4.5 Laplace’s equation in polar coordinates
We now show that a solution u(x, y) of Laplace’s equation in Cartesian coordinates corresponds to a
solution u(r cos θ, r sin θ) of Laplace’s equation in polar coordinates.
Lemma 4.27. Let u(x, y) be a solution of Laplace’s equation
∂ 2u ∂ 2 u
+
= 0.
∂x2 ∂y2
Then the function u(r cos θ, r sin θ) 4.6 is a solution of
∂ 2u 1 ∂u 1 ∂ 2u
+
+
=0
∂r2 r ∂r r2 ∂θ 2
which is Laplace’s equation in polar coordinates.
Proof. Let u(x, y) be a solution of Laplace’s equation. If we replace x, y in u(x, y) by
x = r cos θ
y = r sin θ
we obtain the function
u(r cos θ, r sin θ).
(4.129)
We shall apply Lemma 4.26 to the function (4.129). The corresponding functions r, θ of Lemma 4.26
shall be
1
r = (x2 + y 2) 2
y
θ = tan−1
x
From Figure 4.2 we have
(4.130)
(4.131)
cos θ =
adjacent
x
=
hypotenuse (x2 + y 2) 12
sin θ =
opposite
y
=
hypotenuse (x2 + y 2) 12
Substituting (4.130) and (4.131) into (4.129) gives
u(r cos θ, r sin θ)
2
2
= u (x + y )
1
2
x
2
1
2
, (x + y )
(x2 + y 2) 2
1
2
y
1
(x2 + y 2) 2
!
(4.132)
and notice that the function of x and y given in (4.132) is nothing but the solution u(x, y) of
Laplace’s equation
!
1
1
x
y
2
2 2
2
2 2
= u(x, y).
u (x + y )
1 , (x + y )
1
(x2 + y 2) 2
(x2 + y 2) 2
Applying Lemma 4.26 to (4.132) gives
∂u ∂u ∂r ∂u ∂θ
=
·
+
·
∂x ∂r ∂x ∂θ ∂x
−y
∂u
x
∂u
=
·
+
·
(x2 + y 2)
∂r (x2 + y 2) 12 ∂θ
where r is as given in (4.130).
⇒
∂u ∂u x ∂u
=
· +
·
∂x ∂r r ∂θ
−y
r2
4.6. The functions u(x, y) and u(r cos θ, r sin θ) are the same function expressed in different coordinates.
(4.133)
174
Partial Differential Equations
Also
∂u ∂u ∂r ∂u ∂θ
=
·
+
·
∂y ∂r ∂y ∂θ ∂y
x
∂u
y
∂u
=
·
+
·
(x2 + y 2)
∂r (x2 + y 2) 12 ∂θ
⇒
∂u ∂u y ∂u x =
· +
·
∂y ∂r r ∂θ r2
(4.134)
Notice that equations (4.133) and (4.134) express the partial derivatives of the solution u(x, y) in terms of the partial derivatives of u(r cos θ, r sin θ). Similarly, we want
to express the second partial derivatives
∂ 2u ∂ 2u
,
∂x2 ∂y 2
of the solution u(x, y) in terms of the second partial derivatives of u(r cos θ, r sin θ).
Then the equation
∂ 2u ∂ 2 u
+
=0
∂x2 ∂y2
will give a relationship among the second partial derivatives of u(r cos θ, r sin θ), this
relationship will be exactly Laplace’s equation in polar coordinates
∂ 2u 1 ∂u 1 ∂ 2u
+
+
=0
∂r2 r ∂r r2 ∂θ 2
and the fact that the partial derivatives of u(r cos θ, r sin θ) satisfy this relationship
means that u(r cos θ, r sin θ) is a solution of Laplace’s equation in polar coordinates.
So we differentiate (4.133) with respect to x, by first applying the product rule and then chain rule
in the form of Lemma 4.26
∂ ∂u
∂ 2u
−y
x ∂u ∂ x ∂ ∂u
∂u ∂ − y
=
+
·
+
·
·
·
+
∂x ∂θ
r ∂r ∂x r
∂θ ∂x r 2
∂x2 ∂x ∂r
r2
2
∂ ∂u
−y
x ∂u y
∂u 2xy
∂ ∂u
· +
·
· +
·
+
=
r ∂r r3 ∂x ∂θ
∂θ r4
∂x ∂r
r2
2
−y
x ∂u y 2
∂u 2xy
∂ u x
−y
∂ 2u x ∂ 2u
−y
∂ 2u
·
·
+
=
·
+
·
+
·
+
·
+
·
·
2
3
2
2
2
2
r ∂r r
∂θ r4
∂r r ∂θ∂r
∂r∂θ r ∂θ
r
r
r
⇒
∂ 2u ∂ 2u x2
∂ 2u 2xy ∂ 2u y 2 ∂u y 2 ∂u 2xy
=
·
−
·
+ 2· 4+
· +
·
∂θ r
∂r r3 ∂θ r4
∂x2 ∂r2 r2 ∂θ∂r r3
(4.135)
Similarly, by differentiating (4.134) with respect to y gives
∂ 2u 2xy ∂ 2u x2 ∂u x2 ∂u 2xy
∂ 2u ∂ 2u y 2
=
·
+
·
+ 2· 4+
· −
·
∂θ r
∂r r3 ∂θ r4
∂y 2 ∂r2 r 2 ∂θ∂r r3
by adding (4.135) and (4.136) we have
2
2
2
∂ 2u ∂ 2 u
x + y 2 ∂ 2u
x + y 2 ∂u
x + y 2 ∂ 2u
+
=
+
+
.
r2
r3
r4
∂r 2
∂r
∂θ 2
∂x2 ∂y 2
(4.136)
(4.137)
The function u(x, y) on the left of (4.137) which is a solution of Laplace’s equation by hypothesis
∂ 2u ∂ 2 u
+
=0
∂x2 ∂y2
and substituting this in (4.137) and using the equation
r2 = x2 + y 2
175
4.5 Laplace’s equation in polar coordinates
gives the following relationship between the partial derivatives of u(r cos θ, r sin θ)
∂ 2u 1 ∂u 1 ∂ 2u
+
+
=0
∂r2 r ∂r r2 ∂θ 2
which implies that u(r cos θ, r sin θ) is a solution of Laplace’s equation in polar coordinates.
We illustrate Laplace’s equation in polar coordinates by the following example.
Example. By solving Laplace’s equation in polar coordinates
∂ 2u 1 ∂u 1 ∂ 2u
+
+
=0
∂r2 r ∂r r2 ∂θ2
in the circular domain
(4.138)
{(r, θ)|r 6 1}
find the steady-state temperature of a thin metal disc of radius 1 which has boundary temperature
specified by
u(1, θ) = sin3θ
0 6 θ 6 2π.
(4.139)
Answer: The following diagram illustrates the boundary condition around the metal disc
We use the method of seperation of variables and therefore assume that our solution u(r, θ) takes
the form
u(r, θ) = R(r)Θ(θ).
(4.140)
Substituting the solution (4.84) into Laplace’s equation in polar coordinates (4.138) gives
∂2
1 ∂
1 ∂2
(R(r)Θ(θ)) +
(R(r)Θ(θ)) + 2 2 (R(r)Θ(θ)) = 0
2
r ∂r
r ∂θ
∂r
1
1
⇒ R ′′(r)Θ(θ) + R ′(r)Θ(θ) + 2 R(r)Θ ′′(θ) = 0
r
r
⇒
(r2 R ′′(r) + rR ′(r))Θ(θ) = − R(r)Θ ′′(θ)
⇒
r 2 R ′′(r) + rR ′(r)
Θ ′′(θ)
=−
R(r)
Θ(θ)
and the only way for equation (4.85) to hold is if
r 2 R ′′(r) + rR ′(r)
and
R(r)
functions and are equal to the same constant k. So we have
r2 R ′′(r) + rR ′(r)
Θ ′′(θ)
=−
=k
R(r)
Θ(θ)
(4.141)
−
Θ ′′(θ)
are both constant
Θ(θ)
(4.142)
176
Partial Differential Equations
where the constant k can be zero, negative or positive, that is
k = 0, − a2 or a2
where a > 0.
Now notice that
(r, θ), (r, θ + 2π), (r, θ + 4π), specifiy exactly the same point in polar coordinates. It follows that the function u(r, θ) should have
the following property
u(r, θ) = u(r, θ + 2kπ)
where k is an integer. Therefore from equation (4.140) we have
R(r)Θ(θ) = R(r)Θ(θ + 2kπ)
⇒
Θ(θ) = Θ(θ + 2kπ)
that is, the function Θ(θ) must be periodic with period 2π. This fact implies that the constant k of
equation (4.142) cannot be negative, that is k − a2, because
−
Θ ′′(θ)
= − a2
Θ(θ)
⇒ Θ ′′(θ) − a2 Θ(θ) = 0
⇒ Θ(θ) = Ae−aθ + Beaθ
and this function is only periodic if A = B = 0.
We now consider the other two possibilities k = 0 and k = a2 .
k=0
If k = 0 then from equation (4.142) we obtain two ordinary differential equations
r2 R ′′(r) + rR ′(r) = 0
Θ ′′(θ) = 0
Solving these
4.7
gives
R(r) = A ′ + B ′ ln r
(4.143)
Θ(θ) = C ′ + D ′θ
(4.144)
where A ′, B ′,C ′ and D ′ are constants. Now from above, the function Θ(θ) must be periodic, and
therefore D ′ = 0, hence
Θ(θ) = C ′
(4.145)
(note that a constant function is periodic).
Substituting equations (4.143) and (4.145) into equation (4.140) gives
u(r, θ) = a + b ln r
(4.146)
where a = A ′C ′ and b = B ′C ′.
Now for this steady-state heat problem, we expect that the temperature at r = 0 (that is at the
center of the disc) to have some finite value as the condition
u(1, θ) = sin3θ
0 6 θ 6 2π
4.7. We can solve r2 R ′′(r) + rR ′(r) = 0 by using the substitution w = R ′(r) and then seperating variables.
177
4.5 Laplace’s equation in polar coordinates
imposes finite temperature on the boundary. Note however, that if b 0 in the solution (4.146) then
lim a + b ln r = ± ∞
r→0+
depending on the sign of b because
lim ln r = − ∞.
r→0+
Because we expect finite temperature at r = 0, we have that b = 0 in (4.146) and therefore
u(r, θ) = a
is the only possible solution of Laplace’s equation in polar coordinates when the constant k = 0 in
equation (4.142). Because we shall eventually use Fourier series, we write this solution as
u(r, θ) =
a0
.
2
(4.147)
k = a2
If k = a2 then from equation (4.142) we obtain two ordinary differential equations
r2 R ′′(r) + rR ′(r) − a2R(r) = 0
(4.148)
Θ ′′(θ) + a2Θ(θ) = 0
(4.149)
The solution of (4.149) is
Θ(θ) = A ′′cos(aθ) + B ′′sin(aθ).
and because Θ(θ) must be periodic with period 2π, it follows that
a = n = 1, 2, 3, Substituting a = n into (4.148) gives
r2 R ′′(r) + rR ′(r) − n2R(r) = 0
and this differential equation is called Euler’s equation. By substitution it is easy to check that
the general solution of Euler’s equation is
R(r) = C ′′rn + D ′′r −n
and so, by substituting in (4.140), for each value of n = 1, 2, 3, we have a corresponding solution
un(r, θ) = (Cn′′rn + Dn′′r −n)(An′′cos(nθ) + Bn′′sin(nθ))
(4.150)
As discussed above, we expect that the temperature at r = 0 (that is at the center of the disc) to
have some finite value. Note however, that
lim r −n = ∞,
r→0+
this implies that for each n = 1, 2, 3, we have that the constant
Dn′′ = 0
and so for each value of n = 1, 2, 3, we have the corresponding solution
un(r, θ) = rn(ancos(nθ) + bnsin(nθ))
where an = Cn′′An′′ and bn = Cn′′Bn′′.
So we see that for k = 0 and k = 1, 2, the functions
a0
,
r(a1cos(θ) + b1sin(θ)), r2(a2cos(2θ) + b2sin(2θ)),
2
178
Partial Differential Equations
are each solutions of this particular Laplace’s equation in polar coordinates. So from the Superposition Principle we have that
u(r, θ) =
∞
a0 X n
+
r (ancos(nθ) + bnsin(nθ))
2
(4.151)
n=1
is the form of the solution of our Laplace’s equation in polar coordinates. We determine the constants an and bn of (4.151) from the given boundary condition
u(1, θ) = sin3θ
when 0 6 θ 6 2π.
Substituting r = 1 into (4.151) implies that
∞
a0 X
+
(ancos(nθ) + bnsin(nθ)) = sin3θ
2
when 0 6 θ 6 2π.
(4.152)
n=1
The left side of (4.152) is nothing but the Fourier series of sin3θ for the interval 0 6 θ 6 2π. Therefore, from the Euler’s formulae for a Fourier series we have
Z
1 π
sin3θcos(nθ) dθ
where n = 0, 1, 2, 3, an =
π −π
Z
1 π
bn =
sin3θ sin(nθ)dθ
where n = 1, 2, 3, π −π
Notice that the function
sin3θ
is an odd function, therefore we have that
an = 0
for each n = 0, 1, 2, 3, To determine the bn we recall the triple angle formula
sin 3θ = 3sinθ − 4sin3θ
3
1
⇒ sin3θ = sinθ − sin 3θ
4
4
Also recall that if n is an integer, then
Z π
Z π
sin(nθ)sin (nθ) dθ =
sin2(nθ) dθ
−π
−π
Z π
1 − cos(2nθ)
=
dθ
2
−π
=π
and when n m are both integers
Z π
Z
sin(nθ )sin (mθ) dθ =
−π
= 0.
π
−π
cos(m − n)θ − cos(m + n)θ
dθ
2
Therefore we have
Z
1 π
b1 =
sin3θ sin θ dθ
π Z−π
1
1 π 3
( sinθ − sin 3θ) sin θ dθ
=
4
π Z−π 4
Z
1 π 3
1 π 1
=
sinθ sin θ dθ −
sin 3θ sin θ dθ
π −π 4
π −π 4
1
1 3
= . π − .0
π
π 4
3
=
4
(4.153)
179
4.5 Laplace’s equation in polar coordinates
and similarly
Z
1 π
sin3θ sin 3θ dθ
π Z−π
1
1 π 3
( sinθ − sin 3θ) sin 3θ dθ
=
4
π −π 4
Z
Z
1 π 3
1 π 1
=
sinθ sin 3θ dθ −
sin 3θ sin3θ dθ
π −π 4
π −π 4
1
1 1
= .0 − . π
π
π 4
1
=− .
4
b3 =
For any positive integer n 1, 3 we have
Z
1 π
bn =
sin3θ sin nθ dθ
π Z−π
1 π 3
1
=
( sinθ − sin 3θ) sin nθ dθ
π −π 4
4
Z π
Z
3
1 π 1
1
sinθ sin nθ dθ −
sin 3θ sin nθ dθ
=
π −π 4
π −π 4
1
1
= .0 − .0 (from the identity (4.153))
π
π
=0
and substituting
an = 0
bn =
for each n = 0, 1, 2, 3, 
3

n=1

 4
1
−4



0
n=3
otherwise
into (4.151) we obtain the solution of Laplace’s equation in polar coordinates
u(r, θ) =
1
3
r sinθ − r3 sin 3θ
4
4
that satisfies the given boundary condition (4.139).
180
Partial Differential Equations