Articolo G.A. Partial Differential Equations and Boundary Value Problems with Maple V
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8 Chapter 0
Two vectors ϕ
n
(x) and ϕ
m
(x), which are indexed by the positive integers n and m, are
orthonormal with respect to the weight function w(x) over the interval [a, b] if the following
relation holds:
b
a
ϕ
n
(x)ϕ
m
(x) w(x) dx = δ(n, m)
Here, δ(n, m) is the familiar Kronecker delta function whose value is 0 if n = m and is 1 if
n = m.
Orthonormal sets play a big role in the development of solutions to partial differential
equations.
0.3 Preparation for Ordinary Differential Equations
An ordinary linear homogeneous differential equation of the second order has the form
a2(x)
d
2
dx
2
y(x)
+a1(x)
d
dx
y(x)
+a0(x) y(x) = 0
Here, the coefficients a2(x), a1(x), and a0(x) are functions of the single independent variable
x, and y is the dependent variable of the differential equation. We say the differential equation
is “normal” over some finite interval I if the leading coefficient a2(x) is never zero over that
interval.
Recall that the second derivative of a function is a measure of its concavity, the first derivative
is a measure of its slope, and the zero derivative is a measure of its magnitude. Thus, the
solution y(x) to the above second-order differential equation is that function whose concavity
multiplied by a2(x), plus the slope multiplied by a1(x), plus the magnitude multiplied by a0(x)
must all add up to zero. Finding solutions to such differential equations is standard material for
a course in differential equations.
For now, we state some fundamental theorems about the solution space of ordinary differential
equations.
Theorem 0.1: On any interval I, over which the nth-order linear ordinary homogeneous
differential is normal, the solution space is of finite dimension n and there exist n linearly
independent solution vectors y1(x), y2(x), y3(x),...,yn(x).
Theorem 0.2: If y1(x) and y2(x) are two solutions to a linear second-order differential
equation over some interval I, and the Wronskian of these two solutions does not equal zero
anywhere over this interval, then the two solutions are linearly independent and form a set of
“basis” vectors.
Basic Review 9
From differential equations, the second-order Wronskian of the two vectors y1(x) and y2(x) is
defined as
W(y1(x), y2(x)) = y1(x)
d
dx
y2(x)
−y2(x)
d
dx
y1(x)
Similar to what we do in linear algebra, with a set of basis vectors in hand, we can span the
solution space and write any solution vector as a linear combination of these basis vectors.
As a simple example, one that is analogous to the two-dimensional Euclidean space R2, we
consider the solution space of the linear second-order homogeneous differential equation
d
2
dx
2
y(x) +y(x) = 0
The preceding differential equation is referred to as an “Euler”-type differential equation.
As can easily be verified, two solution vectors are the Euler functions y1(x) = cos(x) and
y2(x) = sin(x). Are the vectors linearly independent? If we evaluate the Wronskian of this set,
we get
W(y1(x), y2(x)) = cos(x)
2
+sin(x)
2
Since the Wronskian is never equal to zero and the differential equation is normal everywhere,
the two vectors form a basis for the solution space of the differential equation. Thus, the set
S1 ={cos(x), sin(x)}
is a basis for the solution space of this particular differential equation.
In terms of this basis, we can span the solution space and write the general solution to the
preceding differential equation as
y(x) = C1 cos(x) +C2 sin(x)
where C1 and C2 are arbitrary constants.
It can be verified that another equivalent basis set is
S2 ={e
ix
, e
−ix
}
Similar to the Euclidean spaces discussed earlier, the two sets S1 and S2 contain two vectors
that are linearly independent; however, the sets themselves are linearly dependent. This follows
from the familiar Euler formulas
sin(x) =
e
ix
−e
−ix
2 i
10 Chapter 0
and
cos(x) =
e
ix
+e
−ix
2
With a set of basis vectors in hand, we can write any solution to a linear differential equation as
a linear superposition of these basis vectors.
0.4 Preparation for Partial Differential Equations
Partial differential equations differ from ordinary differential equations in that the equation has
a single dependent variable and more than one independent variable. We focus on three main
types of partial differential equations in this text, all linear.
1. The heat or diffusion equation (first-order derivative in time t, second-order derivative in
distance x)
∂
∂t
u(x, t) = k
∂
2
∂x
2
u(x, t)
2. The wave equation (second-order derivative in time t, second-order derivative in
distance x)
∂
2
∂t
2
u(x, t) = c
2
∂
2
∂x
2
u(x, t)
3. The Laplace equation (second-order derivative in both distance variables x and y)
∂
2
∂x
2
u(x, y) +
∂
2
∂y
2
u(x, y) = 0
We note that in all three cases, we have a single dependent variable u and more than one
independent variable. The terms c and k are constants.
For the particular types of partial differential equations we will be looking at, all are
characterized by a linear operator, and all of them are solved by the method of separation of
variables. A dramatic difference between ordinary and partial differential equations is the
dimension of the solution space. For ordinary differential equations, the dimension of the
solution space is finite; it is equal to the order of the differential equation. For partial
differential equations with spatial boundary conditions, the dimension of the solution space is
infinite. Thus, a basis for the solution space of a partial differential equation consists of an
infinite number of vectors. As an example, consider the diffusion equation
∂
∂t
u(x, t) = k
∂
2
∂x
2
u(x, t)
Basic Review 11
subject to a given set of spatial boundary conditions. By separation of variables, we assume a
solution in the form of a product
u(x, t) = X(x) T(t)
After substitution of the assumed solution into the partial differential equation, we end up with
two ordinary differential equations: one whose independent variable is x and one whose
independent variable is t.
From the imposition of the given spatial boundary conditions, we find an infinite number of
x-dependent solutions that take on the form of eigenfunctions that are indexed by positive
integers n and written as
X
n
(x)
for n = 0, 1, 2, 3,... .
Similarly, the t-dependent solution can also be indexed by the integer n, and we write the
t-dependent solution as
T
n
(t)
Thus, for a given value of n, one solution to the homogeneous partial differential equation,
which satisfies the boundary conditions, is given as
u
n
(x, t) = X
n
(x) T
n
(t)
for n = 0, 1, 2, 3,... .
Since the partial differential equation operator is linear, any superposition of solutions for all
allowed values of n satisfies the partial differential equation and the given boundary
conditions. Thus, the set of vectors
S ={u
n
(x, t)}
for n = 0, 1, 2, 3,... , forms a basis for the solution space of the partial differential equation.
Since there are an infinite number of indexed solutions, we say the basis of the solution space is
“infinite.” Similar to what we do for ordinary differential equations, we can write the general
solution to the problem as a superposition of the allowed basis vectors—that is,
u(x, t) =
∞
n=0
u
n
(x, t)
The following chapters provide the steps for solving partial differential equations with
boundary conditions.
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CHAPTER 1
Ordinary Linear Differential
Equations
1.1 Introduction
We discuss ordinary linear differential equations in general by initially focusing on the
second-order equation. We begin by considering a normal, linear, second-order,
nonhomogeneous differential equation on some interval I.
a2(t)
d
2
dt
2
y(t)
+a1(t)
d
dt
y(t)
+a0(t)y(t) = f(t)
By “normal” we mean that the leading coefficient a2(t) is not equal to zero anywhere on the
interval I. The coefficients a2(t), a1(t), and a0(t) are, in general, functions of the independent
variable t. The solution y(t) denotes the single dependent variable y to be a function of the
single independent variable t.
The function f(t) is generally referred to as the “driving” or external “source” function. If we
set f(t) = 0, we get the corresponding homogeneous differential equation
a2(t)
d
2
dt
2
y(t)
+a1(t)
d
dt
y(t)
+a0(t)y(t) = 0
Basis Vectors
From theorems covered in Chapter 0, the dimension of the solution space of an ordinary linear
differential equation is equal to the order of the differential equation. Thus, for a second-order
equation, the dimension of the solution space is two, and a set of basis vectors of the system
consists of two linearly independent solutions to the corresponding homogeneous differential
equation.
We define the linear differential equation operator L acting on y(t) as
L(y) = a2(t)
d
2
dt
2
y(t)
+a1(t)
d
dt
y(t)
+a0(t)y(t)
© 2009, 2003, 1999 Elsevier Inc. 13
14 Chapter 1
A basis of the system consists of two solution vectors y1(t) and y2(t), which are linearly
independent and each of which satisfies the corresponding homogeneous equations L(y1) = 0
and L(y2) = 0. There are an infinite number of legitimate sets of basis vectors of the system.
However, all of the sets can be shown to be linearly dependent on each other.
The test for linear independence of the two vectors y1(t) and y2(t) on the interval I is that
the Wronskian W(y1(t), y2(t)) does not equal zero at any point on the interval. Recall, the
Wronskian W(y1(t), y2(t)) is given as
W(y1(t), y2(t)) = y1(t)
d
dt
y2(t)
−y2(t)
d
dt
y1(t)
If y1(t) and y2(t) are solutions of the homogeneous differential equation—that is, L(y1) = 0
and L(y2) = 0—and their Wronskian W(y1(t), y2(t)) does not vanish at any point on the
interval, then these two vectors form a basis of the system on that interval.
If L is normal on the interval I and if y1(t) and y2(t) are the basis vectors of the system, then we
say these two vectors form a “complete” set. This is equivalent to saying that they completely
“span” the solution space to the homogeneous differential equation. Thus, the general solution
to the homogeneous equation y
h
(t) is given as the linear superposition of the basis vectors
y
h
(t) = C1 y1(t) +C2 y2(t)
In the preceding, C1 and C2 are arbitrary constants.
If we denote a particular solution to the nonhomogeneous differential equation as y
p
(t), then
the general solution to the nonhomogeneous equation can be written as a sum of the preceding
homogeneous solution plus the particular solution
y(t) = y
h
(t) +y
p
(t)
We will eventually show that a particular solution to the corresponding nonhomogeneous
differential equation can be constructed from a set of basis vectors of the system.
1.2 First-Order Linear Differential Equations
We consider the first-order linear nonhomogeneous differential equation that is normal on an
interval I and that has the form
a1(t)
d
dt
y(t)
+a0(t) y(t) = f(t)
The corresponding first-order homogeneous equation can be written as
a1(t)
d
dt
y(t)
+a0(t) y(t) = 0
Ordinar y Linear Differential Equations 15
The single basis vector solution to this homogeneous differential equation is
y1(t) = e
−
a0(t)
a1(t)
dt
DEMONSTRATION: We seek the basis vector solution to the first-order linear homogeneous
differential equation
t
2
d
dt
y(t)
+3ty(t) = 0
SOLUTION: We identify the coefficients of the differential equation to be a1(t) = t
2
and
a0(t) = 3t. Note that the differential equation is not normal at the origin. The single system
basis vector is given by the integral
y1(t) = e
−
3
t
dt
This evaluates to be
y1(t) =
1
t
3
for t = 0. We can check this answer by substituting it back into the differential equation.
We now focus on the solution to the corresponding nonhomogeneous differential equation. We
assume the solution to the nonhomogeneous differential equation to be a nonlinear multiple of
this basis vector, and we set
y(t) = u(t)y1(t)
Here, u(t) is some unknown, variable function of t [we are forcing a condition of linear
independence between y(t) and y1(t)]. Substituting this assumed solution into the differential
equation, we get
a1(t)
d
dt
u(t)
y1(t) +a1(t)u(t)
d
dt
y1(t)
+a0(t)u(t)y1(t) = f(t)
Recognizing that y1(t) is a solution to the homogeneous differential equation—that is,
L(y1) = 0—the last two preceding terms vanish, and we get a first-order equation in u(t) that
reads
d
dt
u(t) =
f(t)
a1(t) y1(t)
A simple integration of the preceding yields
u(t) =
f(t)
a1(t) y1(t)
dt
16 Chapter 1
Thus, a particular solution to the nonhomogeneous differential equation can be written as the
product of u(t) and y1(t) shown next:
y
p
(t) = y1(t)
f(t)
a1(t) y1(t)
dt
If we define the first-order Green’s function G1(t, s) as
G1(t, s) =
y1(t)
a1(s) y1(s)
then we can express our particular solution in terms of the preceding Green’s function as
y
p
(t) =
G1(t, s)f(s) ds (1.1)
Writing the solution in terms of the Green’s function illustrates its general usefulness in that
once the Green’s function for a particular differential equation is evaluated, the solution can
accommodate any type of driving or source function f(t).
DEMONSTRATION: We seek the solution to the first-order linear nonhomogeneous
differential equation
t
2
d
dt
y(t)
+3 ty(t) = f(t)
SOLUTION: From the preceding, we identify a1(t) = t
2
and a0(t) = 3 t. The single basis
vector is
y1(t) = e
−
a0(t)
a1(t)
dt
which evaluates to
y1(t) =
1
t
3
From knowledge of the basis vector, we constuct the first-order Green’s function
G1(t, s) =
y1(t)
a1(s)y1(s)
which evaluates to
G1(t, s) =
s
t
3
Ordinar y Linear Differential Equations 17
The particular solution is the integral whose integrand is the product of the Green’s function
and the source function written in terms of the dummy variable s:
y
p
(t) =
sf ( s )
t
3
ds
Thus, the general solution is
y(t) =
C1
t
3
+
sf ( s )
t
3
ds (1.2)
for t = 0. Here, C1 is an arbitrary constant.
Care must be used in evaluating the preceding integral to ensure that we eventually get an
answer that is dependent only on t. The correct procedure is to first perform the integration
with respect to the dummy variable s and then substitute t back in for s. This is illustrated in the
worked-out examples that follow.
As the preceding solution stands, we can appreciate its generality in that it can accommodate
any source function f(t). This is in dramatic contrast with the cumbersome method of
undetermined coefficients whereby one makes a trial-and-error guess at the solution.
We have completed the development of the solution to all first-order linear differential
equations. These solutions will be useful later in the development of solutions to partial
differential equations. We now consider some example problems using Maple.
EXAMPLE 1.2.1: We seek a solution to the first-order linear nonhomogeneous differential
equation
d
dt
y(t) +y(t) = e
t
SOLUTION: We identify the coefficients and the driving function
> restart: a1(t):=1;a0(t):=1;
a1(t) := 1
a0(t) := 1 (1.3)
> f(t):=exp(t);f(s):=subs(t=s,f(t)):
f(t) := e
t
(1.4)
System basis vector
> y1(t):=exp(int(−a0(t)/a1(t),t));
y1(t) := e
−t
(1.5)