ch03.pptx

Differential Equations
James R. Brannan and William E. Boyce
Copyright © 2015 by John Wiley & Sons, Inc.
All rights reserved.
Chapter 03:
Systems of Two
First Order
Equations

Chapter 3 Systems of Two First
Order Equations
 We introduce systems of two first order equations.
 In this chapter, we consider only systems of two first order equations
and we focus most of our attention on systems of the simplest kind: two
first order linear equations with constant coefficients.
 Our goals are to show what kinds of solutions such a system may have
and how the solutions can be determined and displayed graphically, so
that they can be easily visualized.

Chapter 3 Systems of Two First
Order Equations
 3.1 Systems of Two Linear Algebraic Equations
 3.2 Systems of Two First Order Linear Differential Equations
 3.3 Homogeneous Linear Systems with Constant Coefficients
 3.4 Complex Eigenvalues
 3.5 Repeated Eigenvalues
 3.6 A Brief Introduction to Nonlinear Systems

Review linear algebraic systems: Consider the system
a11x1+ a12x2=b1
a21x1+ a22x2=b2
In matrix notation, Ax=b, where
Here, A is a given 2x2 matrix, b a given 2x1 column vector, and x a 2x1
column vector to be determined.
11 12
21 22
a a
A
a a
 
  
 
1
2
x
x
x
 
  
 
1
2
b
b
b
 
  
 
3.1 Systems of Two Linear Algebraic Equations

Solutions to a system of equations
 There are three distinct possibilities for two straight lines in a plane:
they may intersect at a single point, they may be parallel and
nonintersecting, or they may be coincident.
Examples:
 1. 3x1 − x2 = 8, x1 + 2x2 = 5.
 2. x1 + 2x2 = 1, x1 + 2x2 = 5.
 3. 2x1 + 4x2 = 10, x1 + 2x2 = 5.

Cramer’s Rule – THEOREM 3.1.1
The system
a11x1 + a12x2 = b1,
a21x1 + a22x2 = b2,
has a unique solution if and only if the determinant Δ = a11a22 − a12a21 ≠ 0.
The solution is given by
If Δ = 0, then the system has either no solution or infinitely many.

Matrix Method
 Consider coefficient matrix,
 If A−1 exists, then A is called nonsingular or invertible. On the other
hand, if A−1 does not exist, then A is said to be singular or
noninvertible.
 The solution to Ax=B is x = A−1b.







22
21
12
11
a
a
a
a
A







2
1
b
b
b

Homogeneous System
THEOREM 3.1.2
The homogeneous system Ax = 0 always has the trivial solution x1 = 0,
x2 = 0, and this is the only solution when det(A) ≠ 0. Nontrivial solutions
exist if and only if det(A) =0. In this case, unless A = 0, all solutions are
proportional to any nontrivial solution; in other words, they lie on a line
through the origin. If A = 0, then every point in the x1x2-plane is a
solution of system.
Example: Solve the system
3x1 − x2 = 0, x1 + 2x2 = 0.

Eigenvalues and Eigenvectors
 Eigenvalues (λ) of the matrix A are the solutions to Ax = λx. The
eigenvector x corresponding to the eigenvalue λ is obtained by solving
Ax = λx for x for the given λ.
 For a 2x2 matrix Ax = λx reduces to
Since det(A-λI)=0, get
0
22
21
12
11











x
a
a
a
a



Characteristic Equation
The characteristic equation of the matrix A is
λ2 − (a11 + a22)λ + a11a22 − a12a21 = 0.
Solutions determine the eigenvalues.
The two solutions, the eigenvalues λ1 and λ2, may be real and different,
real and equal, or complex conjugates.

Examples
Find the eigenvalues and eigenvectors of the matrix A.
 1.
 2.
 3.

THEOREM 3.1.3
 Let A have real or complex eigenvalues λ1 and λ2 such that λ1≠λ2,
and let the corresponding eigenvectors be x1 and x2. If X is the
matrix with first and second columns taken to be x1 and x2,
respectively, then det(X) ≠0.
That is,

3.2 Systems of Two First Order Linear Differential
Equations
Motivation

Consider the schematic diagram of the greenhouse/rockbed system in Figure 3.2.1.
The rockbed, consisting of rocks ranging in size from 2 to 15 cm, is loosely packed
so that air can easily pass through the void space between the rocks. The rockbed,
and the underground portion of the air ducts used to circulate air through the
system, are thermally insulated from the surrounding soil. Rocks are a good
material for storing heat since they have a high energy-storage capacity, are
inexpensive, and have a long life.
During the day, air in the greenhouse is heated primarily by solar radiation.
Whenever the air temperature in the greenhouse exceeds an upper threshold
value, a thermostatically controlled fan circulates the air through the system,
thereby transferring heat to the rockbed. At night, when the air temperature in the
greenhouse drops below a lower threshold value, the fan again turns on, and heat
stored in the rockbed warms the circulating air.
Example (A Rockbed Heat Storage System)

We wish to study temperature variation in the greenhouse during the nighttime
phase of the cycle. A simplified model for the system is provided by lumped
system thermal analysis, in which we treat the physical system as if it consists of
two interacting components.
 Assume that the air in the system is well mixed so that both the temperature of
the air in the greenhouse and the temperature of the rockbed are functions of
time, but not location. Let us denote the air temperature in the greenhouse by
𝑢1(𝑡) and the temperature of the rockbed by 𝑢2(𝑡). We will measure t in hours
and temperature in degrees Celsius.
 The following table lists the relevant parameters that appear in the mathematical
model below. We use the subscripts 1 and 2 to indicate thermal and physical
properties of the air and the rock medium, respectively.

The units of 𝐶1, and 𝐶2 are J/kg⋅◦C, while the units of ℎ1 and ℎ2 are J/h⋅m2⋅◦C. The
area of the air-rock interface is approximately equal to the sum of the surface areas
of the rocks in the rock storage bed.
Using the law of conservation of energy

The units of 𝐶1, and 𝐶2 are J/kg⋅◦C, while the units of ℎ1 and ℎ2 are J/h⋅m2⋅◦C. The
area of the air-rock interface is approximately equal to the sum of the surface areas
of the rocks in the rock storage bed.
Using the law of conservation of energy, we get the differential equation

 Equation (1) states that the rate of change of energy in the air internal to the
system equals the rate at which heat flows across the above-ground enclosure
(made of glass or polyethylene) plus the rate at which heat flows across the
underground air-rock interface. In each case, the rates are proportional to the
difference in temperatures of the materials on each side of the interface. The
algebraic signs that multiply each term on the right are chosen so that heat flows in
the direction from hot to cool.
 Equation (2) arises from the following reasoning. Since the rockbed is insulated
around its boundary, heat can enter or leave the rockbed only across the air-rock
interface. Energy conservation requires that the rate at which heat is gained or lost
by the rockbed through this interface must equal the heat lost or gained by the
greenhouse air through the same interface. Thus the right side of Eq. (2) is equal to
the negative of the second term on the right-hand side of Eq. (1).

Matrix Notation, Vector Solutions, and Component
Plots
du/dt = Ku + b.
where K is a given 2X2 matrix and b a given 2x1 matrix. U is 2X1
matrix of unknowns whose first derivative is du/dt. We solve this
system subject to a given initial condition u(0)= u0, a 2X1 matrix with
given values.

Terminology
 The components of u are scalar valued functions of t, so we can plot
their graphs. Plots of u1 and u2 versus t are called component plots.
 The variables u1 and u2 are often called state variables, since their
values at any time describe the state of the system.
 Similarly, the vector u = u1i + u2j is called the state vector of the
system. The u1u2-plane itself is called the state space. If there are
only two state variables, the u1u2-plane may be called the state plane
or, more commonly, the phase plane.

Geometry of Solutions: Direction Fields
The right side of a system of first order equations du/dt = Ku + b
defines a vector field that governs the direction and speed of
motion of the solution at each point in the phase plane.
Because the vectors generated by a vector field for a specific
system often vary significantly in length, it is customary to scale
each nonzero vector so that they all have the same length. These
vectors are then referred to as direction field vectors for the
system and the resulting picture is called the direction field.

Geometry of Solutions: Phase Portraits
Using a computer we can to generate solution trajectories. A
plot of a representative sample of the trajectories, including
any constant solutions, is called a phase portrait of the
system of equations.

Phase portrait for the example

Solutions of Two First Order Linear Equations
THEOREM 3.2.1 Existence and Uniqueness of Solutions
Let each of the functions 𝑝11, . . . , 𝑝22, 𝑔1, and 𝑔2 be continuous on
an open interval 𝐼 = 𝛼 < 𝑡 < 𝛽, let 𝑡0 be any point in 𝐼, and let 𝑥0
and 𝑦0 be any given numbers. Then there exists a unique
solution of the system
that also satisfies the initial conditions
Further, the solution exists throughout the interval 𝐼.

First Order Linear Equations
 In matrix form, we write
 The system above is called a first order linear system of
dimension two because it consists of first order equations and
because its state space (the 𝑥𝑦-plane) is two-dimensional.
 Further, if 𝑔(𝑡) = 0 for all 𝑡, that is, 𝑔1(𝑡) = 𝑔2(𝑡) = 0 for all 𝑡,
then the system is said to be homogeneous. Otherwise, it is
nonhomogeneous.

If the right side of
does not depend explicitly on the independent variable t, the system is
said to be autonomous.
Linear Autonomous Systems
Then the coefficient matrix P and the components of the vector g must
be constants. We use the notation dx/dt = Ax + b, where A is a
constant matrix and b is a constant vector, to denote autonomous
linear systems.

Critical points of linear autonomous system
For the linear autonomous system, we find the equilibrium
solutions, or critical points, by setting dx/dt equal to zero.
Hence any solution of Ax=−b is a critical point of the system.
If the coefficient matrix A has an inverse, as we usually assume, then
Ax = −b has a single solution, namely, x=−A−1b. This is then the only
critical point of the system.
However, if A is singular, then Ax = −b has either no solution or
infinitely many

Transformation of a Second Order Equation to a
System of First Order Equations
Consider the second order equation
y'' + p(t)y' + q(t)y = g(t),
where p, q, and g are given functions that we assume to be continuous
on an interval I.
Substituting x1 = y and x2 = y'. This system can be transformed to a
system of two first order equations,

Example
Consider the differential equation
u'' + 0.25u' + 2u = 3 sin t.
Suppose that initial conditions u(0) = 2, u’(0) = −2. Transform this
problem into an equivalent one for a system of first order equations.
Write the matrix notation for this initial value problem.
Answer:

Component plots of the solution to this initial value
problem

3.3 Homogeneous Linear Systems with Constant
Coefficients
Reducing x' = Ax + b to x' = Ax
If A has an inverse, then the only critical, or equilibrium, point of
x' = Ax + b is xeq = −A−1b
In such cases it is convenient to shift the origin of the phase plane to the
critical point using the coordinate transformation x = xeq + x˜.
Substituting, we get dx˜/dt = Ax˜. Therefore, if x = φ(t) is a solution of
the homogeneous system x' = Ax, then the solution of the
nonhomogeneous system x' = Ax + b is given by
x = φ(t) + xeq = φ(t) − A−1b.

The Eigenvalue Method for Solving x' = Ax
Consider a general system of two first order linear homogeneous
differential equations with constant coefficients dx/dt=Ax.
x = eλtv is a solution of dx/dt = Ax provided that λ is an eigenvalue and
v is a corresponding eigenvector of the coefficient matrix A.
Hence Av = λv, or (A − λI)v = 0.

THEOREM 3.3.1 - Principle of Superposition
If we assume that λ1 and λ2 are eigenvalues (real and different of A) we
have:
Suppose that x1(t) = eλ1𝑡v1 and x2(t) = eλ2𝑡 v2 are solutions of
dx/dt = Ax.
Then the expression
x = c1x1(t) + c2x2(t),
where c1 and c2 are arbitrary constants, is also a solution.

Example
Consider the system dx/dt = x.
Find solutions of the system and then find the particular solution that
satisfies the initial Condition
x(0) = .
Answer:










4
0
0
1








3
2

















 

1
0
3
0
1
2 4t
t
e
e
x

Wronskian Determinant
The determinant
is called the Wronskian determinant or, more simply, the Wronskian of
the two vectors x1 and x2. If x1(t) = eλ1tv1 and x2(t) = eλ2tv2, then their
Wronskian is
Two solutions x1(t) and x2(t) of whose Wronskian is not zero are referred
to as a fundamental set of solutions. The linear combination of x1
and x2 given with arbitrary coefficients c1 and c2, x = c1x1(t) + c2x2(t), is
called the general solution.

THEOREM 3.3.3
Suppose that x1(t) and x2(t) are two solutions of dx/dt = Ax, and that their
Wronskian is not zero. Then x1(t) and x2(t) form a fundamental set of
solutions, and the general solution is given by, x = c1x1(t) + c2x2(t),
where c1 and c2 are arbitrary constants. If there is a given initial
condition x(t0) = x0, where x0 is any constant vector, then this condition
determines the constants c1 and c2 uniquely.
Note: The theorem is true if coefficient matrix A has eigenvalues that are
real and different. It is also valid even when the eigenvalues are
complex or repeated.

EXAMPLE - A Rockbed Heat Storage System
Consider again the greenhouse/rockbed heat storage problem with
coordinates centered at the critical point given by,
dx/dt = x = Ax.
Find the general solution of this system. Then plot a direction field, a
phase portrait, and several component plots of the system.














4
1
4
1
4
3
8
13

Answer
 The general solution is
 The eigenvalues are λ1 = -7/4 and λ2 = −1/8. Direction field and
phase portrait for the system is shown in the next slide.

Nodal Sources and Nodal Sinks
The pattern of trajectories in
Figure is typical of all second
order systems x' = Ax whose
eigenvalues are real, different,
and of the same sign. The origin
is called a node for such a
system.

Nodal Sources and Nodal Sinks
 If the eigenvalues were positive rather than negative, then the
trajectories would be similar but traversed in the outward direction.
 Nodes are asymptotically stable if the eigenvalues are negative and
unstable if the eigenvalues are positive.
 Asymptotically stable nodes and unstable nodes are also referred to
as nodal sinks and nodal sources respectively.

Example
Consider the system
dx/dt = x = Ax.
Find the general solution and draw a phase portrait.








1
4
1
1

Answer
 The general solution is
 The eigenvalues are λ1 = 3 and λ2 = −1. Direction field and phase
portrait for the system is shown in the next slide.

Saddle Points
The pattern of trajectories in
Figure is typical of all second
order systems x' = Ax for which
the eigenvalues are real and of
opposite signs. The origin is
called a saddle point in this
case.
Saddle points are always
unstable because almost all
trajectories depart from them as t
increases.

Phase portraits for x′ = Ax when A has distinct
real eigenvalues

3.4 Complex Eigenvalues
Consider a two dimensional system x= Ax with complex conjugate
eigenvalues
To solve the system, find the eigenvalues and eigenvectors, observing
that they are complex conjugates. Then write down x1(t) and separate
it into its real and imaginary parts u(t) and w(t), respectively. Finally,
form a linear combination of u(t) and w(t), x = c1u(t) + c2w(t).
Of course, if complex-valued solutions are acceptable, you can simply
use the solutions x1(t) and x2(t).
Thus Theorem 3.3.3 is also valid when the eigenvalues are complex.

Procedure for Finding the General Solution
(Complex Eigenvalues)

Example
Consider the system
Find a fundamental set of solutions and display them graphically in a
phase portrait and component plots.
Answer: The General solution

Component plots for the solutions u(t) and w(t) of
the system

A direction field and phase portrait for the system

Spiral Points
The phase portrait in previous Figure is typical of all two-dimensional
systems x' = Ax whose eigenvalues are complex with a negative real part.
The origin is called a spiral point and is asymptotically stable because all
trajectories approach it as t increases. Such a spiral point is often called a
spiral sink.
For a system whose eigenvalues have a positive real part, the trajectories
are similar to those in Figure, but the direction of motion is away from the
origin and the trajectories become unbounded. In this case, the origin is
unstable and is often called a spiral source.

Centers
If the real part of the eigenvalues is zero, then
there is no exponential factor in the solution
and the trajectories neither approach the
origin nor become unbounded. Instead, they
repeatedly traverse a closed curve about the
origin.
An example of this behavior can be seen in
Figure to left. In this case, the origin is called
a center and is said to be stable, but not
asymptotically stable. In all three cases, the
direction of motion may be either clockwise,
as in previous Example, or counterclockwise,
depending on the elements of the coefficient
matrix A.

Summary
For two-dimensional systems with real coefficients, we have now
completed our description of the three main cases that can occur:
1. Eigenvalues are real and have opposite signs; x = 0 is a saddle point.
2. Eigenvalues are real and have the same sign but are unequal; x = 0 is
a node.
3. Eigenvalues are complex with nonzero real part; x = 0 is a spiral point.

Phase portraits for x′ = Ax when A has complex
eigenvalues

3.4 Repeated Eigenvalues. An Example
Consider the system x' = Ax, where
Draw a direction field, a phase portrait, and typical component plots.
Answer: The eigenvalues are λ1 = λ2 = −1. General solution
x(t) = c1x1(t) + c2x2(t)
where

Typical component plots for the system

Proper Node or Star Point
 It is possible to show that the only 2x2 matrices with a repeated
eigenvalue and two independent eigenvectors are the diagonal
matrices with the eigenvalues along the diagonal. Such matrices form
a rather special class, since each of them is proportional to the
identity matrix. The system in above Example is entirely typical of this
class of systems.
 In this case the origin is called a proper node or, sometimes, a star
point.

Repeated Eigenvalues (in general)
Consider two-dimensional linear homogeneous systems with constant
coefficients given by x' = Ax.
Suppose that λ1 is a repeated eigenvalue of the matrix A and that there is
only one independent eigenvector v1.
 Then one solution is x1(t) = e λ1t v1.
 A second solution is x2(t) = teλ1tv1 + eλ1tw, where w satisfies
(A − λ1I)w = v1.

Repeated Eigenvalues (Cont.)
 The vector w is called a generalized eigenvector corresponding to
the eigenvalue λ1.
 In the case where the 2x2 matrix A has a repeated eigenvalue and
only one eigenvector, the origin is called an improper or degenerate
node.

Example
Consider the system
Find the eigenvalues and eigenvectors of the coefficient matrix, and then
find the general solution of the system. Draw a direction field, phase
portrait, and component plots.
Answer: The eigenvalues are λ1 = λ2 = −1/2. General solution
x(t) = c1x1(t) + c2x2(t)
where

A direction field and phase portrait for the system
improper or
degenerate node

Typical plots of x1 versus t for the system

Phase portraits for x′ = Ax when A has a single
repeated eigenvalue

Stability diagram
𝑝 and 𝑞 are the trace and determinant, respectively, of the coefficient matrix
A of the given system

3.6 A Brief Introduction to Nonlinear Systems
 In Section 3.2, we introduced the general two-dimensional first order
linear system
(1)
(2)
 Of course, two-dimensional systems that are not of the form (1) or
(2) may also occur. Such systems are said to be nonlinear.

THEOREM 3.6.1 - Existence and Uniqueness of
Solutions
Let each of the functions 𝑓 and 𝑔 and the partial derivatives 𝜕𝑓/𝜕𝑥,
𝜕𝑓/𝜕𝑦, 𝜕𝑔/𝜕𝑥, and 𝜕𝑔/𝜕𝑦 be continuous in a region 𝑅 of 𝑡𝑥𝑦-space
defined by 𝛼 < 𝑡 < 𝛽, 𝑎 < 𝑥 < 𝑏, 𝑐 < 𝑦 < 𝑑, and let the point (𝑡0, 𝑥0, 𝑦0)
be in 𝑅. Then there is an interval |𝑡 − 𝑡0| < ℎ in which there exists a
unique solution of the system of differential equations
that also satisfies the initial conditions
𝑥(𝑡0) = 𝑥0, 𝑦(𝑡0) = 𝑦0.

Autonomous Systems
 It is usually impossible to solve nonlinear systems exactly by analytical
methods.
 Therefore for such systems graphical methods and numerical
approximations become even more important. Here we will consider
systems for which direction fields and phase portraits are of particular
importance. These are systems that do not depend explicitly on the
independent variable t. In other words, the functions f and g in the
equation depend only on x and y and not on t.
 Such a system is called autonomous, and can be written in the form

Equilibrium Points or Critical Points
 To find equilibrium, or constant, solutions of the autonomous system, we
set 𝑑𝑥/𝑑𝑡 and 𝑑𝑦/𝑑𝑡 equal to zero, and solve the resulting equations
𝑓 𝑥, 𝑦 = 0, 𝑔(𝑥, 𝑦) = 0
for 𝑥 and 𝑦. Any solution of these is a point in the phase plane that is a
trajectory of an equilibrium solution. Such points are called equilibrium
points or critical points.
 Depending on the particular forms of 𝑓 and 𝑔, the nonlinear system can
have any number of critical points, ranging from none to infinitely many.

Example
 Consider the system
𝑑𝑥
𝑑𝑡
= 𝑥 − 𝑦
𝑑𝑦
𝑑𝑡
= 2𝑥 − 𝑦 − 𝑥2
 Find a function 𝐻(𝑥, 𝑦) such that the trajectories of the system lie on the
level curves of 𝐻. Find the critical points and draw a phase portrait for
the given system. Describe the behavior of its trajectories.

Example (Cont.)
 To find the critical points, solve the equations
𝑥 − 𝑦 = 0, 2𝑥 − 𝑦 − 𝑥2 = 0.
 The critical points are (0, 0) and (1, 1).
 To determine the trajectories, note that for this system, becomes
𝑑𝑦
𝑑𝑥
=
2𝑥 − 𝑦 − 𝑥2
𝑥 − 𝑦
 This is exact and so solutions satisfy
𝐻(𝑥, 𝑦) = 𝑥2 − 𝑥𝑦 +
1
2
𝑦2 −
1
3
𝑥3 = 𝑐,
where c is an arbitrary constant.

A phase portrait for the system

Section 3.1 Two-Dimensional Linear Algebra

Section 3.2 Systems of Two First Order Linear
Equations

Section 3.3 Homogeneous Systems with Constant
Coefficients: x' = Ax

Section 3.4 Complex Eigenvalues

Section 3.5 Repeated Eigenvalues

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