Chapter 8.1
Preliminary theory – Linear system
In this section we will study how to solve systems of first-order linear differential
equations of the form:
dx1
 a11 (t )x1  a12 (t )x2 
dt
dx2
 a21 (t )x1  a22 (t )x2 
dt
 a1n (t )xn  f1 (t )
dxn
 an1 (t )x1  an2 (t )x2 
dt
 ann (t )xn  fn (t )
 a2 n (t )xn  f2 (t )
The system above is called the normal form.
When fi (t )  0 for i  1,, n , then the system is said to be homogeneous.
Otherwise, it is nonhomogeneous.
We assume that the coefficients aij as well as the functions fi are continuous on
a common interval I .
Let
 x1 (t ) 
 x (t ) 
X  2  ,




 xn (t ) 
 a11 (t ) a12 (t )
 a (t ) a (t )
22
A(t )   21


 an1 (t ) an2 (t )
a1n (t ) 
a2n (t ) 
,


ann (t ) 
 f1 (t ) 
 f (t ) 
F (t )   2 




 fn (t ) 
Then the system of linear first-order differential equations can be written as
 x1   a11 (t ) a12 (t )
  
d  x2   a21 (t ) a22 (t )

dt   
  
 xn   an1 (t ) an2 (t )
a1n (t )  x1   f1 (t ) 
  f (t ) 
a2 n (t ) 
x
2
    2 
  

  

ann (t )  xn   fn (t ) 
or simply X '  AX  F . If the system is homogeneous, its matrix form is X '  AX .
Definition: (Solution vector)
A solution vector on an interval I is any column matrix
 x1 (t ) 
 x (t ) 
X  2 




 xn (t ) 
whose entries are differentiable functions satisfying the system above on the
interval.
Theorem: (Superposition principle)
Let X1 , X2 ,, X k be a set of solution vectors of the homogeneous system X '  AX
on an interval I . Then the linear combination
X  c1 X1  c2 X2 
 ck X k
where ci , i  1,2,, k are arbitrary constants, is also a solution of the system of
the interval.
Definition: (Fundamental set of solutions)
Any set X1 , X2 ,, X n of n linearly independent solution vectors of the
homogeneous system X '  AX on an interval I is said to be a fundamental set of
solutions on the interval.
Theorem: (General solution – Homogeneous systems)
Let X1 , X2 ,, X n be a fundamental set of solutions of the homogeneous system
X '  AX on an interval I . Then the general solution of the system on the interval
is
X  c1 X1  c2 X2 
where ci , i  1,2,, n are arbitrary constants.
 cn X n
Theorem: (General solution – Nonhomogeneous systems)
Let X p be a given solution of the nonhomogeneous system X '  AX  F on an
interval and let
X c  c1 X1  c2 X2 
 cn X n
denote the general solution on the same interval of the associated homogeneous
system X '  AX . Then the general solution of the nonhomogeneous system on
the interval is X  X c  X p .
Write the linear system in matrix form.
dx
t
Example 1 (6): dt  3x  4 y  e sin2t
dy
 5x  9 z  4e t cos 2t
dt
dz
 y  6 z  e t
dt
Verify that the vector X is a solution of the linear system.
dx
Example 2 (12): dt  2 x  5y
 5 cos t
 t
dy
 2 x  4 y
X 
e
dt
 3cos t  sint 
Homework 8.1: 1-15 odd
Chapter 8.2
Homogeneous linear systems
Definition: (Eigenvalue and eigenvector)
Let A be an n  n matrix (i.e. a linear transformation from
 is an eigenvalue of A and that K  0 ( K 
n
n
to
n
). We say that
) is an associated eigenvector if
and only if
AK  K
How to find eigenvalues:
We are looking for K and  such that
AK  K  AK  K  0  (A  I)K  0
1
0
Recall: I  


0
0
1
0
We want this system to have a nontrivial solution ( x  0 ).
Therefore, it is necessary for the matrix A  I to be singular, i.e., the
determinant of A  I has to be equal to 0 ( det(A  I)  0 is called the
characteristic equation).
Once all the values of  are known, we can use (A  I)K  0 to find the
corresponding eigenvectors K .
0


0

1
Theorem: (General solution – Homogeneous systems)
Let 1 , 2 ,, n be n DISTINCT real eigenvalues of the coefficient matrix A of the
homogeneous system X '  AX and let K1 , K2 ,, K n be the corresponding
eigenvectors. Then the general solution of the system X '  AX on the interval
(, ) is given by
X  c1K1e1t  c2K2e2t 
 cnKnent
Find the general solution of the given system.
Example 1 (6):
dx
 6 x  2y
dt
dy
 3x  y
dt
Why do we use eigenvalues and eigenvectors?
Given the system X '  AX , let K be any vector and  be any number.
We try as a solution X  Ke t . Then X '  K  e t .
Plug it in the equation:
K et  AKet  K   AK   is an eigenvalue of A with eigenvector K .
We are now interested on finding the solutions of the system when some of the
eigenvalues are repeated. We will have two different cases.
Eigenvalue of multiplicity m. Case 1.
For some n  n matrices A it may be possible to find m linearly independent
vectors K1 , K2 ,, K m corresponding to an eigenvalue 1 of multiplicity m  n . In
this case the solution of the system contains the linear combination
c1K1e1t  c2K2e1t 
 cmKme1t
Eigenvalue of multiplicity m. Case 2.
If there is only one eigenvector corresponding to the eigenvalue 1 of multiplicity
m , then linearly independent solutions of the form
X1  K11e 1t
X2  K21te 1t  K22e 1t
t m1 1t
t m2 1t
X m  K m1
e  K m2
e 
(m  1)!
(m  2)!
 K mme 1t
where K ij are column vectors, can always be found.
Obs.: Case 2 is an oversimplification. Given an eigenvalue of multiplicity m , there
are up to m corresponding independent eigenvectors. For example, a certain
matrix may have an eigenvalue of multiplicity 4 and there exist 2 corresponding
independent eigenvectors. (Not just one, as case 2 states)
Example 2 (26):
dx
 3 x  2y  4 z
dt
dy
 2 x  2z
dt
dz
 4 x  2y  3z
dt
Example 3 (26):
dx
 6 x  5y
dt
dy
 5x  4 y
dt
Repeated eigenvalues (justification)
Given the system X '  AX , assume  is an eigenvalue with algebraic multiplicity 2
but only one eigenvector K .
We have one solution X1  Ket .
We try as a second solution X2  tX1  Pet  Ktet  Pet .
Then X2 '  K (et  tet )  Pet  K tet  (K  P )et .
Plug in X2 into X '  AX :
X2 '  AX2  K tet  (K  P )et  AKtet  APet
Dividing by e t :
K t  K  P  AKt  AP
Thus, equating terms of same degree:
1) K   AK (We knew this already: K is an eigenvalue of A with eigenvalue  )
2) K  P  AP  AP  P  K 
(A  I)P  K
Complex eigenvalue.
Let A be the coefficient matrix having real entries of the homogeneous system
X '  AX , and let K1 be an eigenvector corresponding to the complex eigenvalue
1    i  , with  and  real. Let Q1 be the conjugate of K1 and 1 be the
conjugate of 1 . Then
K1e1t
and
Q1e 1t
are solutions of X '  AX .
Obs.: In order to obtain a solution with real basis vectors 
Let 1     i and K 1  A  Bi . Then the solutions are:
X 1  et (A cos  t  B sin  t )
X 2  et (B cos  t  A sin  t )
Example 4 (26): dx  4 x  5y
dt
dy
 2 x  6y
dt
Homework 8.2: 1-13 odd, 21-29 odd, 35-45 odd
Chapter 8.3
Nonhomogeneous linear systems
As we learned in chapter 4, we can find particular solutions of nonhomogeneous
linear differential equations by the methods of undetermined coefficients (4.4) or
variation of parameters (4.6).
The method of undetermined coefficients, although sometimes efficient, does not
adapt very well to the solution of nonhomogeneous system of equations.
Therefore, we will concentrate on variation of parameters, which is the more
powerful technique.
Recall the following theorem from section 8.1:
Theorem: (General solution – Nonhomogeneous systems)
Let X p be a given solution of the nonhomogeneous system X '  AX  F on an
interval and let
X c  c1 X1  c2 X2 
 cn X n
denote the general solution on the same interval of the associated homogeneous
system X '  AX . Then the general solution of the nonhomogeneous system on
the interval is X  X c  X p .
Use variation of parameters to solve the given nonhomogeneous system.
Example 1 (14):
dx
 2x  y
dt
dy
 3x  2y  4t
dt
Variation of parameters (justification)
Let    X1
Xn  , where X i , i  1,, n are column vectors that are
X2
solutions of the homogeneous system X '  AX . Then  '  A .
Note that  1 exists because X1 , X2 ,, X n are a fundamental set of solutions
(linearly independent) and, therefore, det   0 .
 u1 (t ) 
 with u (t),,u (t) functions.
We try X p  U , where U  
1
n

 u (t ) 
 n 
Using the product rule to differentiate, we have X p '   'U  U ' . (Note that the
order of the factors is crucial for the product to be defined).
Plug in X '  AX  F :
 'U  U '  A(U)  F
Since  '  A ,
AU  U '  AU  F  U '  F  U '   1F  U    1F dt
Example 2 (22):
Example 3 (28):
dx
 3 x  2y  1
dt
dy
 2 x  y  1
dt
 0 1
 1 
X'
X


 cot t 
 1 0 


Homework 8.3: 13-29 odd
Chapter 10.1 Autonomous systems
Definition: (Autonomous systems)
A system of first-order differential equations is said to be autonomous when the
system can be written in the form
dx1
 g1 (x1 , x2 , , xn )
dt
dx2
 g2 (x1 , x2 , , xn )
dt
dxn
 gn (x1 , x2 ,
dt
, xn )
Observe that the independent variable t does not appear explicitly on the righthand side of each differential equation.
Obs. 1: If n  1 then we have a separable equation.
Obs. 2: Any second-order autonomous differential equation can be written as an
x  y
d2x
autonomous system. Example: 2  3cos x  0 can be written as
y  3cos x  0
dt
Obs. 3: The autonomous system in the definition can be expressed in column
vector form as X   g( X ) (as in the previous chapter) where
 x1 (t ) 
 x (t ) 
X (t )   2 




 xn (t ) 
and
 g1 (x1 ,, xn ) 
 g (x ,, x ) 
n 
g(X)   2 1




 gn (x1 ,, xn ) 
However, in this chapter it will also be convenient to use row vector notation
where
X (t)   x1 (t),, xn (t)
and
g(X)   g1 (x1 ,, xn ),, gn (x1 ,, xn )
When n  2 the system is called a plane autonomous system and we can write
the system as
dx
 P( x , y )
dt
dy
 Q(x , y)
dt
Under specific conditions on the continuity of P , Q , and its partial derivatives,
then a solution of the plane autonomous system that satisfies the initial condition
X (0)  X 0 is unique and one of three basic types.
a) A constant solution of the form x(t )  x0 , y(t)  y0 . A solution of this form is
called a critical or stationary point. When the particle is placed at the
critical point, it remains there indefinitely (that is why such solution is also
called equilibrium solution). Such systems are easy to recognize because
x(t )  0, y(t )  0 and, therefore, the plane system is
P( x , y )  0
Q(x , y)  0
b) A solution x  x(t ), y  y(t ) that defines an arc (a plane curve that does not
cross itself)
c) A periodic solution (also called cycle). If p is the period of the curve, then a
particle starting at (x0 , y0 ) will cycle around the curve and return to (x0 , y0 )
after p units of time.
Write the nonlinear second-order differential equation as a plane autonomous
system. Find all critical points.
Example 1 (6): x  x  x x  0 , for  0
Find all critical points.
Example 2 (16): x   x(4  y 2 )
y  4 y(1  x 2 )
Homework 10.1: 1-15 odd
Chapter 10.2 Stability of linear systems
In this section we will study the behavior of solutions of a plane autonomous
system when the initial condition X (0)  X 0 is chosen “close” to a critical point X 1
of the system. In this section X 1  (0,0) .
We are interested in the following two questions:
1) Will the particle return to the critical point? More precisely, lim X (t )  X1 ?
t 
2) If the particle does not return to the critical point, does it remain close to
the critical point or move away from it?
Given the plane autonomous system
x  ax  by
y  cx  dy
the eigenvalues of the system are given by

   2  4
2
where   a  d (the trace of the coefficient matrix) and   ad  bc (the
determinant of the coefficient matrix).
The behavior of the solutions can be discussed through examination of the
eigenvalues.
Case I: two real distinct eigenvalues ( 2  4   0 )
Let 1 and 2 be the eigenvalues. Then
X (t )  c1 K 1e 1t  c2 K 2e 2t  e 1t (c1 K 1  c2 K 2e(2 1 )t )
a) 2  1  0 (Stable node) Note that lim e1t (c1 K 1  c2 K 2e(2 1 )t )  0 and the
t 
solution curves will approach 0 .
If c1  0 , then X (t ) approaches 0 along K1
If c1  0 , then X (t ) approaches 0 along K2
b) 0  2  1 (Unstable node) Note that lim e1t (c1 K 1  c2 K 2e(2 1 )t )  "  " and the
t 
solution curves will move away from 0 .
If c1  0 , then X (t ) moves away from 0 along K1
If c1  0 , then X (t ) moves away from 0 along K2
c) 2  0  1 (Saddle point) Note that lim e1t (c1 K 1  c2 K 2e(2 1 )t )  "  " and some
t 
solution curves will move away from 0 and other solutions will approach 0 .
If c1  0 , then X (t ) moves away from 0 along K1
If c1  0 , then X (t ) approaches 0 along K2
The general solution of the linear system X   AX is given. Discuss the nature of
solutions in a neighborhood of (0,0) .
 1 2 
Example 1 (2): A  
,
3 4
 1
 4 
X (t )  c1   et  c2   e2t
 1 
 6
Case II: two repeated real eigenvalues ( 2  4   0 )
a) Two linearly independent eigenvectors If K1 and K2 are linearly independent
eigenvectors corresponding to 1 , then X (t )  c1 K 1e 1t  c2 K 2e 1t  e 1t (c1 K 1  c2 K 2 )
(Degenerate stable node) If 1  0 then lim X (t )  0 and X (t ) approaches 0
t 
along the line determined by the vector c1 K 1  c2 K 2 .
(Degenerate unstable node) If 1  0 then
lim X (t )  "  " and X (t ) moves away from 0 along
t 
the line determined by the vector c1 K 1  c2 K 2 .
(Same picture with arrows reversed)
b) One linearly independent eigenvector
c
c 

Now X (t )  c1 K 1e 1t  c2 (K 1te 1t  Pe 1t )  te 1t  c2 K 1  1 K 1  2 P 
t
t 

(Degenerate stable node) If 1  0 then lim X (t )  0 and X (t ) approaches 0
t 
along K 1 .
(Degenerate unstable node) If 1  0 then lim X (t )  "  "
t 
and X (t ) moves aways from 0 along K 1 . (reverse graph)
Classify the critical point (0,0) of the given linear system.
Example 2 (14):
3
1
x  x  y
2
4
1
y   x  y
2
Case III: complex eigenvalues ( 2  4   0 )
Let 1     i and 2     i . Then, using section 8.2,
x(t )  et (c11 cos  t  c12 sin  t),
y(t )  et (c21 cos  t  c22 sin  t )
a)   0 (equivalently,   0 )
Then x(t ), y(t ) are periodic with period 2 /  and
all solution curves are ellipses with center at the
critical point (0,0) , called center.
b)   0 (equivalently,   0 ) All solution curves are spirals.
If   0 the spirals get closer and closer to the origin and the critical point (0,0)
is called a stable spiral point.
If   0 the spirals get farther and farther from the origin and the critical point
(0,0) is called an unstable spiral point.
Classify the critical point (0,0) of the given linear system.
Example 3 (12): x  5x  3y
y  7 x  4 y
Homework 10.2: 1-15 odd
Chapter 10.3 Linearization and local stability
Recall from Calculus I:
Given a function f (x) and a point (x1 , f (x1 )) on the graph of the function, then the
tangent line to f at the point (x1 , f (x1 )) is:
y  f (x1 )  f (x1 )(x  x1 )
For values of x close to x1 , the values of y obtained from the equation of the
tangent line are close to the values of y obtained from f .
Analogously, from Calculus III:
Given a function f (x , y) and a point (x1 ,y1 , f (x1 ,y1 )) on the graph of the function,
then the tangent plane to f at the point (x1 ,y1 , f (x1 ,y1 )) is:
z  f (x1 , y1 )  fx (x1 ,y1 )(x  x1 )  fy (x1 ,y1 )(y  y1 )
For values of (x , y) close to (x1 , y1 ) , the values of z obtained from the equation of
the tangent plane are close to the values of z obtained from f .
These approximations are called linearizations. We aim to use linearization to
study nonlinear differential equations and systems.
Stable and unstable critical points
Stable critical point: Let X1 be a critical point of an autonomous system. Given any
disk of radius  with center at X1 , there is always a solution X (t ) inside the disk
provided that the initial condition X (0)  X 0 is sufficiently close to X1 . Then X1 is
called an asymptotically stable critical point.
Graphically:
Unstable critical point: Let X1 be a critical point of an autonomous system. There
is a disk of radius  and center at X1 , such that there exists at least a solution
X (t ) that will leave the disk for some value of t , no matter how close the initial
condition X (0)  X 0 is to X1 . Then X1 is called an unstable critical point.
Graphically:
Linearization: It is many times hard (or impossible) to determine the stability of a
critical point on a nonlinear system by finding explicit solutions. Instead of
working with the original nonlinear system, we will use a process called
linearization.
a) For first order differential equations of the form x  g(x) (note that
x  x(t ) is a function of t ), let x1 be a critical point.
The tangent line to the curve y  g(x) at x  x1 is y  g(x1 )  g '(x1 )(x  x1 ) .
Since x1 is a critical point, g(x1 )  0 and x  g(x)  g '(x1 )(x  x1 ) .
We now concentrate on solving the equation x  g '(x1 )(x  x1 ) .
This is a linear equation as in 2.3, and its solution is x  x1  Ceg( x1 )t .
Stability criteria for x  g(x)
If g(x1 )  0 then lim x(t )  lim (x1  Ceg( x1 )t )  x1 , and then x1 is an
t 
t 
asymptotically stable critical point.
If g(x1 )  0 then lim x(t )  lim (x1  Ceg( x1 )t )   , and then x1 is an unstable
t 
critical point.
t 
Without solving explicitly, classify the critical points of the given first-order
autonomous differential equation as either asymptotically stable or unstable. All
constants are assumed to be positive.
Example 1 (4):
dx
x
 kx ln ,
dt
K
Example 2 (8):
dx
 k(  x)(  x)(  x) ,
dt
x 0
   
b) For plane autonomous systems, we follow a similar approach using the
Jacobian matrix of the system (Calculus III). Details in pages 408-409.
The Jacobian matrix of the system X   g( X ) , more explicitly,
dx
 P( x , y )
dt
dy
 Q(x , y)
dt
evaluated at the critical point X1  (x1 , y1 ) is defined as
 P

 x ( x1 ,y1 )
J( X 1 )  
Q

 x ( x ,y )
1 1


P

y ( x1 ,y1 ) 

Q

y ( x1 ,y1 ) 
Stability criteria for plane autonomous systems
Let X1 be a critical point of the plane autonomous system X   g( X )
If the eigenvalues of J( X1 ) have negative real part, then X1 is an
asymptotically stable critical point.
If J( X1 ) has an eigenvalue with positive real part, then X1 is an unstable
critical point.
Once a critical point X1 of a non-linear autonomous system has been classified as
stable or unstable, it is natural to wonder if we can infer more geometrical
information about the solutions near X1 .
In five separate cases (stable node, stable spiral point, unstable spiral point,
unstable node, and saddle) the solutions have the same general geometric
features as the solutions to the linear system, and the smaller the neighborhood
about X1 , the closer the resemblance. Otherwise, it is not possible to further
classify the geometric nature of the solutions.
Classify (if possible) each critical point of the given plane autonomous system as a
stable node, a stable spiral point, an unstable spiral point, an unstable node, or a
saddle point.
Example 3 (12): x  x 2  y 2  1
y   2y
Example 4 (16): x  xy  3y  4
y  y 2  x 2
Homework 10.3: 3-23 odd