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Boyce/DiPrima/Meade 11th ed, Ch 7.1: Introduction to Systems
of First Order Linear Equations
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• A system of simultaneous first order ordinary differential
equations has the general form
where each xk is a function of t. If each Fk is a linear
function of x1, x2, …, xn, then the system of equations is said
to be linear, otherwise it is nonlinear.
• Systems of higher order differential equations can similarly
be defined.
),,,(
),,,(
),,,(
21
2122
2111
nnn
n
n
xxxtFx
xxxtFx
xxxtFx
Example 1
• The motion of a certain spring-mass system from Section 3.7
was described by the differential equation
• This second order equation can be converted into a system of
first order equations by letting x1 = u and x2 = u'. Thus
or
¢¢u (t)+1
8¢u (t)+ u(t) = 0
¢x1 = x2
¢x2 +1
8x2 + x1 = 0
¢x1 = x2
¢x2 = -x1 -1
8x2
Nth Order ODEs and Linear 1st Order
Systems
• The method illustrated in the previous example can be used
to transform an arbitrary nth order equation
into a system of n first order equations, first by defining
Then
)1()( ,,,,, nn yyyytFy
)1(
321 ,,,, n
n yxyxyxyx
),,,( 21
1
32
21
nn
nn
xxxtFx
xx
xx
xx
Solutions of First Order Systems
• A system of simultaneous first order ordinary differential
equations has the general form
It has a solution on if there exists n functions
that are differentiable on I and satisfy the system of
equations at all points t in I.
• Initial conditions may also be prescribed to give an IVP:
).,,,(
),,,(
21
2111
nnn
n
xxxtFx
xxxtFx
)(,),(),( 2211 txtxtx nn
0
0
0
202
0
101 )(,,)(,)( nn xtxxtxxtx
I :a < t < b
Theorem 7.1.1
• Suppose F1,…, Fn and
are continuous in the region R of t x1 x2…xn-space defined by
and let the point
be contained in R. Then in some interval
(t0 – h, t0 + h) there exists a unique solution
that satisfies the IVP.
00
2
0
10 ,,,, nxxxt
)(,),(),( 2211 txtxtx nn
),,,(
),,,(
),,,(
21
2122
2111
nnn
n
n
xxxtFx
xxxtFx
xxxtFx
¶F1 / ¶x1,...,¶F1 / ¶xn,...,¶Fn / ¶x1,...,¶Fn / ¶xn
a < t < b,a1 < x1 < b1,...,an < xn < bn
Linear Systems
• If each Fk is a linear function of x1, x2, …, xn, then the
system of equations has the general form
• If each of the gk(t) is zero on I, then the system is
homogeneous, otherwise it is nonhomogeneous.
)()()()(
)()()()(
)()()()(
2211
222221212
112121111
tgxtpxtpxtpx
tgxtpxtpxtpx
tgxtpxtpxtpx
nnnnnnn
nn
nn
Theorem 7.1.2
• Suppose p11, p12,…, pnn, g1,…, gn are continuous on an
interval with t0 in I, and let
prescribe the initial conditions. Then there exists a unique
solution
that satisfies the IVP, and exists throughout I.
00
2
0
1 ,,, nxxx
)(,),(),( 2211 txtxtx nn
)()()()(
)()()()(
)()()()(
2211
222221212
112121111
tgxtpxtpxtpx
tgxtpxtpxtpx
tgxtpxtpxtpx
nnnnnnn
nn
nn
I :a < t < b
Boyce/DiPrima/Meade 11th ed, Ch 7.2:
Review of Matrices
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• For theoretical and computational reasons, we review results
of matrix theory in this section and the next.
• A matrix A is an m x n rectangular array of elements,
arranged in m rows and n columns, denoted
• Some examples of 2 x 2 matrices are given below:
mnmm
n
n
ji
aaa
aaa
aaa
a
21
22221
11211
A
ii
iCB
7654
231,
42
31,
43
21A
Transpose
• The transpose of A = (aij) is AT = (aji).
• For example,
mnnn
m
m
T
mnmm
n
n
aaa
aaa
aaa
aaa
aaa
aaa
21
22212
12111
21
22221
11211
AA
63
52
41
654
321,
42
31
43
21TT
BBAA
Conjugate
• The conjugate of A = (aij) is A = (aij).
• For example,
mnmm
n
n
mnmm
n
n
aaa
aaa
aaa
aaa
aaa
aaa
21
22221
11211
21
22221
11211
AA
443
321
443
321
i
i
i
iAA
Adjoint
• The adjoint of A is AT , and is denoted by A*
• For example,
mnnn
m
m
mnmm
n
n
aaa
aaa
aaa
aaa
aaa
aaa
21
22212
12111
*
21
22221
11211
AA
432
431
443
321*
i
i
i
iAA
Square Matrices
• A square matrix A has the same number of rows and
columns. That is, A is n x n. In this case, A is said to have
order n.
• For example,
nnnn
n
n
aaa
aaa
aaa
21
22221
11211
A
987
654
321
,43
21BA
Vectors
• A column vector x is an n x 1 matrix. For example,
• A row vector x is a 1 x n matrix. For example,
• Note here that y = xT, and that in general, if x is a column
vector x, then xT is a row vector.
3
2
1
x
321y
The Zero Matrix
• The zero matrix is defined to be 0 = (0), whose dimensions
depend on the context. For example,
,
00
00
00
,000
000,
00
00
000
Matrix Equality
• Two matrices A = (aij) and B = (bij) are equal if aij = bij for
all i and j. For example,
BABA
43
21,
43
21
Matrix – Scalar Multiplication
• The product of a matrix A = (aij) and a constant k is defined
to be kA = (kaij). For example,
302520
151055
654
321AA
Matrix Addition and Subtraction
• The sum of two m x n matrices A = (aij) and B = (bij) is
defined to be A + B = (aij + bij). For example,
• The difference of two m x n matrices A = (aij) and B = (bij)
is defined to be A - B = (aij - bij). For example,
1210
86
87
65,
43
21BABA
44
44
87
65,
43
21BABA
Matrix Multiplication
• The product of an m x n matrix A = (aij) and an n x r matrix
B = (bij) is defined to be the matrix C = (cij), where
• Examples (note AB does not necessarily equal BA):
n
k
kjikij bac1
417
15
61000512
340023
10
21
03
,654
321
2014
1410
164122
12291
2511
115
16983
8341
42
31,
43
21
CDDC
BA
ABBA
Example 1: Matrix Multiplication
• To illustrate matrix multiplication and show that it is not
commutative, consider the following matrices:
• From the definition of matrix multiplication we have:
112
011
112
,
112
120
121
BA
AB =
2 - 2 + 2 1+ 2 -1 -1+1
2 - 2 -2 +1 -1
4 +1+ 2 2 -1-1 -2 +1
æ
è
çç
ö
ø
÷÷
=
2 2 0
0 -1 -1
7 0 -1
æ
è
çç
ö
ø
÷÷
BA =
2 - 2 -4 + 2 -1 2 -1-1
1 -2 - 2 1+1
2 + 2 -4 - 2 +1 2 +1+1
æ
è
çç
ö
ø
÷÷
=
0 -3 0
1 -4 2
4 -5 4
æ
è
çç
ö
ø
÷÷¹ AB
Vector Multiplication
• The dot product of two n x 1 vectors x & y is defined as
• The inner product of two n x 1 vectors x & y is defined as
• Example:
n
k
ji
T yx1
yx
n
k
ji
T yx,1
yxyx
iiii,
iiii
i
i
i
T
T
2118)55)(3()32)(2()1)(1(
912)55)(3()32)(2()1)(1(
55
32
1
,
3
2
1
yxyx
yxyx
Vector Length
• The length of an n x 1 vector x is defined as
• Note here that we have used the fact that if x = a + bi, then
• Example:
2/1
1
2
2/1
1
2/1||
n
k
k
n
k
kk xxx,xxx
3016941
)43)(43()2)(2()1)(1(
43
2
12/1
ii,
i
xxxx
222 xbabiabiaxx
Orthogonality
• Two n x 1 vectors x & y are orthogonal if (x,y) = 0.
• Example:
0)1)(3()4)(2()11)(1(
1
4
11
3
2
1
yxyx ,
Identity Matrix
• The multiplicative identity matrix I is an n x n matrix
given by
• For any square matrix A, it follows that AI = IA = A.
• The dimensions of I depend on the context. For example,
100
010
001
I
987
654
321
987
654
321
100
010
001
,43
21
10
01
43
21IBAI
Inverse Matrix
• A square matrix A is nonsingular, or invertible, if there
exists a matrix B such that that AB = BA = I. Otherwise A
is singular.
• The matrix B, if it exists, is unique and is denoted by A–1
and is called the inverse of A.
• It turns out that A–1 exists iff detA ≠ 0, and A–1 can be found
using row reduction (also called Gaussian elimination) on
the augmented matrix (A|I), see example on next slide.
• The three elementary row operations:
– Interchange two rows.
– Multiply a row by a nonzero scalar.
– Add a multiple of one row to another row.
Example 2: Finding the Inverse of a Matrix (1 of 2)
• Use row reduction to find the inverse of the matrix A below,
if it exists.
• Solution: If possible, use elementary row operations to
reduce (A|I),
such that the left side is the identity matrix, for then the
right side will be A–1. (See next slide.)
322
213
111
A
,
100322
010213
001111
IA
Example 2: Finding the Inverse of a Matrix (2 of 2)
• Thus
5/15/25/4100
2/12/12/1010
10/310/110/7001
124500
02/12/32/510
02/12/12/301
124500
02/12/32/510
02/12/12/301
102540
02/12/32/510
001111
102540
013520
001111
100322
010213
001111
IA
5/15/25/4
2/12/12/1
10/310/110/71
A
Matrix Functions
• The elements of a matrix can be functions of a real variable.
In this case, we write
• Such a matrix is continuous at a point, or on an interval
(a, b), if each element is continuous there. Similarly with
differentiation and integration:
)()()(
)()()(
)()()(
)(,
)(
)(
)(
)(
21
22221
11211
2
1
tatata
tatata
tatata
t
tx
tx
tx
t
mnmm
n
n
m
Ax
b
aij
b
a
ijdttadtt
dt
da
dt
d)()(, A
A
Example & Differentiation Rules
• Example:
• Many of the rules from calculus apply in this setting. For
example:
dt
d
dt
d
dt
d
dt
d
dt
d
dt
d
dt
d
dt
d
BAB
AAB
BABA
CA
CCA
matrixconstant a is where,
41
0)(
,0sin
cos6
4cos
sin3)(
3
0
2
dtt
t
tt
dt
d
t
ttt
A
AA
Boyce/DiPrima/Meade 11th ed, Ch 7.3: Systems of Linear
Equations, Linear Independence, Eigenvalues
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• A system of n linear equations in n variables,
can be expressed as a matrix equation Ax = b:
• If b = 0, then system is homogeneous; otherwise it is nonhomogeneous.
nnnnnn
n
n
b
b
b
x
x
x
aaa
aaa
aaa
2
1
2
1
,2,1,
,22,21,2
,12,11,1
,,22,11,
2,222,211,2
1,122,111,1
nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
Nonsingular Case
• If the coefficient matrix A is nonsingular, then it is
invertible and we can solve Ax = b as follows:
• This solution is therefore unique. Also, if b = 0, it follows
that the unique solution to Ax = 0 is x = A–10 = 0.
• Thus if A is nonsingular, then the only solution to Ax = 0 is
the trivial solution x = 0.
bAxbAIxbAAxAbAx 1111
Example 1: Nonsingular Case (1 of 3)
• From a previous example, we know that the matrix A below
is nonsingular with inverse as given.
• Using the definition of matrix multiplication, it follows that
the only solution of Ax = 0 is x = 0:
4/14/34/1
4/14/74/5
4/14/54/3
,
112
211
3211
AA
0
0
0
0
0
0
4/14/34/1
4/14/74/5
4/14/54/310Ax
Example 1: Nonsingular Case (2 of 3)
• Now let’s solve the nonhomogeneous linear system Ax = b
below using A–1:
• This system of equations can be written as Ax = b, where
• Then
0834
2301
220
321
321
321
xxx
xxx
xxx
1
1
2
4
5
7
4/14/34/1
4/14/74/5
4/14/54/31bAx
4
5
7
,,
112
211
321
3
2
1
bxA
x
x
x
Example 1: Nonsingular Case (3 of 3)
• Alternatively, we could solve the nonhomogeneous linear
system Ax = b below using row reduction.
• To do so, form the augmented matrix (A|b) and reduce,
using elementary row operations.
1
1
2
1
2
732
1100
2110
7321
4400
2110
7321
10730
2110
7321
10730
2110
7321
4112
5211
7321
3
32
321
x
bA
x
xx
xxx
42
52
732
321
321
321
xxx
xxx
xxx
Singular Case
• If the coefficient matrix A is singular, then A-1 does not
exist, and either a solution to Ax = b does not exist, or there
is more than one solution (not unique).
• Further, the homogeneous system Ax = 0 has more than one
solution. That is, in addition to the trivial solution x = 0,
there are infinitely many nontrivial solutions.
• The nonhomogeneous case Ax = b has no solution unless
(b, y) = 0, for all vectors y satisfying A*y = 0, where A* is
the adjoint of A.
• In this case, Ax = b has solutions (infinitely many), each of
the form x = x(0) + , where x(0) is a particular solution of
Ax = b, and is any solution of Ax = 0.
x
x
Example 2: Singular Case (1 of 2)
• Solve the nonhomogeneous linear system Ax = b below using row
reduction. Observe that the coefficients are nearly the same as in the
previous example
• We will form the augmented matrix (A|b) and use some of the steps in
Example 1 to transform the matrix more quickly
03
30
32
3000
110
321
312
211
321
321
321
2132
1321
321
21
1
3
2
1
bbb
bbb
bbxx
bxxx
bbb
bb
b
b
b
b
bA
3321
2321
1321
32
2
32
bxxx
bxxx
bxxx
Example 2: Singular Case (2 of 2)
• From the previous slide, if , there is no solution
to the system of equations
• Requiring that , assume, for example, that
• Then the reduced augmented matrix (A|b) becomes:
• It can be shown that the second term in x is a solution of the
nonhomogeneous equation and that the first term is the most
general solution of the homogeneous equation, letting ,
where α is arbitrary
0
3
4
1
1
1
3
4
00
3
232
3000
110
321
3
3
3
3
32
321
321
21
1
x
x
x
x
xx
xxx
bbb
bb
b
xx
03 321 bbb
5,1,2 321 bbb
03 321 bbb
3321
2321
1321
32
2
32
bxxx
bxxx
bxxx
3x
Linear Dependence and Independence
• A set of vectors x(1), x(2),…, x(n) is linearly dependent if
there exists scalars c1, c2,…, cn, not all zero, such that
• If the only solution of
is c1= c2 = …= cn = 0, then x(1), x(2),…, x(n) is linearly
independent.
0xxx )()2(
2
)1(
1
n
nccc
0xxx )()2(
2
)1(
1
n
nccc
Example 3: Linear Dependence (1 of 2)
• Determine whether the following vectors are linear
dependent or linearly independent.
• We need to solve
or
0
0
0
1131
112
421
0
0
0
11
1
4
3
1
2
1
2
1
3
2
1
21
c
c
c
ccc
0xxx )3(
3
)2(
2
)1(
1 ccc
11
1
4
,
3
1
2
,
1
2
1)3()2()1(
xxx
Example 3: Linear Dependence (2 of 2)
• We can reduce the augmented matrix (A|b), as before.
• So, the vectors are linearly dependent:
• Alternatively, we could show that the following determinant is zero:
numberany becan where
1
3
2
00
03
042
0000
0310
0421
01550
0930
0421
01131
0112
0421
3332
321
cccc
ccc
c
bA
11
1
4
,
3
1
2
,
1
2
1)3()2()1(
xxx
0
1131
112
421
)det(
ijx
0xxx )3()2()1(
3 32,1if c
Linear Independence and Invertibility
• Consider the previous two examples:
– The first matrix was known to be nonsingular, and its column vectors
were linearly independent.
– The second matrix was known to be singular, and its column vectors
were linearly dependent.
• This is true in general: the columns (or rows) of A are linearly
independent iff A is nonsingular iff A-1 exists.
• Also, A is nonsingular iff detA ≠ 0, hence columns (or rows)
of A are linearly independent iff detA ≠ 0.
• Further, if A = BC, then det(C) = det(A)det(B). Thus if the
columns (or rows) of A and B are linearly independent, then
the columns (or rows) of C are also.
Linear Dependence & Vector Functions
• Now consider vector functions x(1)(t), x(2)(t),…, x(n)(t), where
• As before, x(1)(t), x(2)(t),…, x(n)(t) is linearly dependent on I if
there exists scalars c1, c2,…, cn, not all zero, such that
• Otherwise x(1)(t), x(2)(t),…, x(n)(t) is linearly independent on I
See text for more discussion on this.
,,,,2,1,
)(
)(
)(
)(
)(
)(
2
)(
1
Itnk
tx
tx
tx
t
k
m
k
k
k
x
Ittctctc n
n allfor ,)()()( )()2(
2
)1(
1 0xxx
Eigenvalues and Eigenvectors
• The eqn. Ax = y can be viewed as a linear transformation
that maps (or transforms) x into a new vector y.
• Nonzero vectors x that transform into multiples of
themselves are important in many applications.
• Thus we solve Ax = x or equivalently, (A – I)x = 0.
• This equation has a nonzero solution if we choose such
that det(A – I) = 0.
• Such values of are called eigenvalues of A, and the
nonzero solutions x are called eigenvectors.
l l
ll
l
Example 4: Eigenvalues (1 of 3)
• Find the eigenvalues and eigenvectors of the matrix A.
• Solution: Choose such that det(A – I) = 0, as follows.
24
13A
1,2
122
4123
24
13det
10
01
24
13detdet
2
IA
l l
Example 4: First Eigenvector (2 of 3)
• To find the eigenvectors of the matrix A, we need to solve
(A – I)x = 0 for = 2 and = –1.
• Eigenvector for = 2: Solve
and this implies that . So
0
0
44
11
0
0
224
123
2
1
2
1
x
x
x
x0xIA
1
1 choosearbitrary,
1
1)1(
2
2)1(xx cc
x
x
21 xx
l l l
l
Example 4: Second Eigenvector (3 of 3)
• Eigenvector for = –1: Solve
and this implies that . So
0
0
14
14
0
0
124
113
2
1
2
1
x
x
x
x0xIA
4
1choosearbitrary,
4
1
4
)2(
1
1)2(xx cc
x
x
12 4xx
l
Normalized Eigenvectors
• From the previous example, we see that eigenvectors are
determined up to a nonzero multiplicative constant.
• If this constant is specified in some particular way, then the
eigenvector is said to be normalized.
• For example, eigenvectors are sometimes normalized by
choosing the constant so that ||x|| = (x, x)½ = 1.
Algebraic and Geometric Multiplicity
• In finding the eigenvalues of an n x n matrix A, we solve
det(A – I) = 0.
• Since this involves finding the determinant of an n x n
matrix, the problem reduces to finding roots of an nth
degree polynomial.
• Denote these roots, or eigenvalues, by 1, 2, …, n.
• If an eigenvalue is repeated m times, then its algebraic
multiplicity is m.
• Each eigenvalue has at least one eigenvector, and a
eigenvalue of algebraic multiplicity m may have q linearly
independent eigevectors, 1 ≤ q ≤ m, and q is called the
geometric multiplicity of the eigenvalue.
ll
l l l
Eigenvectors and Linear Independence
• If an eigenvalue has algebraic multiplicity 1, then it is
said to be simple, and the geometric multiplicity is 1 also.
• If each eigenvalue of an n x n matrix A is simple, then A
has n distinct eigenvalues. It can be shown that the n
eigenvectors corresponding to these eigenvalues are linearly
independent.
• If an eigenvalue has one or more repeated eigenvalues, then
there may be fewer than n linearly independent eigenvectors
since for each repeated eigenvalue, we may have q < m.
This may lead to complications in solving systems of
differential equations.
l
Example 5: Eigenvalues (1 of 5)
• Find the eigenvalues and eigenvectors of the matrix A.
• Solution: Choose such that det(A – I) = 0, as follows.
011
101
110
A
1,1,2
)1)(2(
23
11
11
11
detdet
221
2
3
IA
l l
Example 5: First Eigenvector (2 of 5)
• Eigenvector for = 2: Solve (A – I)x = 0, as follows.
1
1
1
choosearbitrary,
1
1
1
00
011
011
0000
0110
0101
0000
0110
0211
0330
0330
0211
0112
0121
0211
0211
0121
0112
)1(
3
3
3
)1(
3
32
31
xx cc
x
x
x
x
xx
xx
l l
Example 5: 2nd and 3rd Eigenvectors (3 of 5)
• Eigenvector for = –1: Solve (A – I)x = 0, as follows.
1
1
0
,
1
0
1
choose
arbitrary,where,
1
0
1
0
1
1
00
00
0111
0000
0000
0111
0111
0111
0111
)3()2(
3232
3
2
32
)2(
3
2
321
xx
x xxxx
x
x
xx
x
x
xxx
l l
Example 5: Eigenvectors of A (4 of 5)
• Thus three eigenvectors of A are
where x(2), x(3) correspond to the double eigenvalue
• It can be shown that x(1), x(2), x(3) are linearly independent.
• Hence A is a 3 x 3 symmetric matrix (A = AT ) with 3 real
eigenvalues and 3 linearly independent eigenvectors.
1
1
0
,
1
0
1
,
1
1
1)3()2()1(
xxx
011
101
110
A
l = -1.
Example 5: Eigenvectors of A (5 of 5)
• Note that we could have we had chosen
• Then the eigenvectors are orthogonal, since
• Thus A is a 3 x 3 symmetric matrix with 3 real eigenvalues
and 3 linearly independent orthogonal eigenvectors.
1
2
1
,
1
0
1
,
1
1
1)3()2()1(
xxx
0,,0,,0, )3()2()3()1()2()1( xxxxxx
Hermitian Matrices
• A self-adjoint, or Hermitian matrix, satisfies A = A*,
where we recall that A* = AT .
• Thus for a Hermitian matrix, aij = aji.
• Note that if A has real entries and is symmetric (see last
example), then A is Hermitian.
• An n x n Hermitian matrix A has the following properties:
– All eigenvalues of A are real.
– There exists a full set of n linearly independent eigenvectors of A.
– If x(1) and x(2) are eigenvectors that correspond to different
eigenvalues of A, then x(1) and x(2) are orthogonal.
– Corresponding to an eigenvalue of algebraic multiplicity m, it is
possible to choose m mutually orthogonal eigenvectors, and hence A
has a full set of n linearly independent orthogonal eigenvectors.
Boyce/DiPrima/Meade 11th ed, Ch 7.4: Basic Theory of Systems
of First Order Linear Equations
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• The general theory of a system of n first order linear equations
parallels that of a single nth order linear equation.
• This system can be written as x' = P(t)x + g(t), where
)()()()(
)()()()(
)()()()(
2211
222221212
112121111
tgxtpxtpxtpx
tgxtpxtpxtpx
tgxtpxtpxtpx
nnnnnnn
nn
nn
)()()(
)()()(
)()()(
)(,
)(
)(
)(
)(,
)(
)(
)(
)(
21
22221
11211
2
1
2
1
tptptp
tptptp
tptptp
t
tg
tg
tg
t
tx
tx
tx
t
nnnn
n
n
nn
Pgx
Vector Solutions of an ODE System
• A vector x = (t) is a solution of x' = P(t)x + g(t) if the components of x,
satisfy the system of equations on .
• For comparison, recall that x' = P(t)x + g(t) represents our system of equations
• Assuming P and g continuous on I, such a solution exists by Theorem 7.1.2.
),(,),(),( 2211 txtxtx nn
)()()()(
)()()()(
)()()()(
2211
222221212
112121111
tgxtpxtpxtpx
tgxtpxtpxtpx
tgxtpxtpxtpx
nnnnnnn
nn
nn
f
I :a < t < b
Homogeneous Case; Vector Function
Notation
• As in Chapters 3 and 4, we first examine the general
homogeneous equation x' = P(t)x.
• Also, the following notation for the vector functions
x(1), x(2),…, x(k),… will be used:
,
)(
)(
)(
)(,,
)(
)(
)(
)(,
)(
)(
)(
)(2
1
)(
2
22
12
)2(
1
21
11
)1(
tx
tx
tx
t
tx
tx
tx
t
tx
tx
tx
t
nn
n
n
k
nn
xxx
Theorem 7.4.1
• If the vector functions x(1) and x(2) are solutions of the system
x' = P(t)x, then the linear combination c1x(1) + c2x
(2) is also a
solution for any constants c1 and c2.
• Note: By repeatedly applying the result of this theorem, it
can be seen that every finite linear combination
of solutions x(1), x(2),…, x(k) is itself a solution to x' = P(t)x.
)()()( )()2(
2
)1(
1 tctctc k
kxxxx
Theorem 7.4.2
• If x(1), x(2),…, x(n) are linearly independent solutions of the
system x' = P(t)x for each point in , then each
solution x = (t) can be expressed uniquely in the form
• If solutions x(1),…, x(n) are linearly independent for each
point in , then they are fundamental solutions
on I, and the general solution is given by
)()()( )()2(
2
)1(
1 tctctc n
nxxxx
)()()( )()2(
2
)1(
1 tctctc n
nxxxx
f
I :a < t < b
I :a < t < b
The Wronskian and Linear Independence
• The proof of Thm 7.4.2 uses the fact that if x(1), x(2),…, x(n)
are linearly independent on I, then detX(t) ≠ 0 on I, where
• The Wronskian of x(1),…, x(n) is defined as
W[x(1),…, x(n)](t) = detX(t).
• It follows that W[x(1),…, x(n)](t) ≠ 0 on I iff x(1),…, x(n) are
linearly independent for each point in I.
,
)()(
)()(
)(
1
111
txtx
txtx
t
nnn
n
X
Theorem 7.4.3
• If x(1), x(2),…, x(n) are solutions of the system x' = P(t)x on
, then the Wronskian W[x(1),…, x(n)](t) is either identically
zero on I or else is never zero on I.
• This result relies on Abel’s formula for the Wronskian
where c is an arbitrary constant (Refer to Section 3.2)
• This result enables us to determine whether a given set of
solutions x(1), x(2),…, x(n) are fundamental solutions by
evaluating W[x(1),…, x(n)](t) at any point t in .
dttptptp
nn
nn
cetWpppdt
dW )]()()([
2211
2211
)()(
I :a < t < b
a < t < b
Theorem 7.4.4
• Let
• Let x(1), x(2),…, x(n) be solutions of the system x' = P(t)x,
, that satisfy the initial conditions
respectively, where t0 is any point in . Then
x(1), x(2),…, x(n) are form a fundamental set of solutions of
x' = P(t)x.
1
0
0
0
,,
0
0
1
0
,
0
0
0
1
)()2()1(
neee
,)(,,)( )(
0
)()1(
0
)1( nn tt exex
a < t < b
a < t < b
Theorem 7.4.5
• Consider the system
where each element of P is a real-valued continuous function. If x =
u(t) + iv(t) is a complex-valued solution of Eq. (3), then its real part
u(t) and its imaginary part v(t) are also solutions of this equation.
x' = P(t)x
Boyce/DiPrima/Meade 11th ed, Ch 7.5: Homogeneous Linear
Systems with Constant Coefficients
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• We consider here a homogeneous system of n first order linear
equations with constant, real coefficients:
• This system can be written as x' = Ax, where
nnnnnn
nn
nn
xaxaxax
xaxaxax
xaxaxax
2211
22221212
12121111
nnnn
n
n
m aaa
aaa
aaa
tx
tx
tx
t
21
22221
11211
2
1
,
)(
)(
)(
)( Ax
Equilibrium Solutions
• Note that if n = 1, then the system reduces to
• Recall that x = 0 is the only equilibrium solution if a ≠ 0.
• Further, x = 0 is an asymptotically stable solution if a < 0,
since other solutions approach x = 0 in this case.
• Also, x = 0 is an unstable solution if a > 0, since other
solutions depart from x = 0 in this case.
• For n > 1, equilibrium solutions are similarly found by
solving Ax = 0. We assume detA ≠ 0, so that x = 0 is the
only solution. Determining whether x = 0 is asymptotically
stable or unstable is an important question here as well.
atetxaxx )(
Phase Plane
• When n = 2, then the system reduces to
• This case can be visualized in the x1x2-plane, which is called
the phase plane.
• In the phase plane, a direction field can be obtained by
evaluating Ax at many points and plotting the resulting
vectors, which will be tangent to solution vectors.
• A plot that shows representative solution trajectories is
called a phase portrait.
• Examples of phase planes, directions fields, and phase
portraits will be given later in this section.
2221212
2121111
xaxax
xaxax
Solving Homogeneous System
• To construct a general solution to x' = Ax, assume a solution
of the form x = ert, where the exponent r and the constant
vector are to be determined.
• Substituting x = ert into x' = Ax, we obtain
• Thus to solve the homogeneous system of differential
equations x' = Ax, we must find the eigenvalues and
eigenvectors of A.
• Therefore x = ert is a solution of x' = Ax provided that r is
an eigenvalue and is an eigenvector of the coefficient
matrix A.
0ξIAAξξAξξ rreer rtrt
x
x
x
x
x
Example 1 (1 of 2)
• Find the general solution of the system
• The most important feature of this system is that the
coefficient matrix is a diagonal matrix. Thus, by writing the
system in scalar form, we obtain
• Each of these equations involves only one of the unknown
variables, so we can solve the two equations separately. In
this way we find that
where c1 and c2 are arbitrary constants.
xx
30
02
x '1 = 2x1, x '2 = -3x2
x1 = c 1e2t , x2 = c2e
-3t
Example 1 (2 of 2)
x =c1e
2t
c2e-et
æ
è
çç
ö
ø
÷÷
= c1
e2t
0
æ
èç
ö
ø÷ + c2
0
e-3t
æ
èç
ö
ø÷ = c1
1
0
æ
èç
ö
ø÷e2t + c2
0
1
æ
èç
ö
ø÷e-3t
• Then by writing the solution in vector form we have
• Now we define two solutions x(1) and x(2) so that
• The Wronskian of these solutions is
which is never zero. Therefore, x(1) and x(2) form a
fundamental set of solutions.
x(1)(t) =1
0
æ
èç
ö
ø÷e2t , x(2)(t) =
0
1
æ
èç
ö
ø÷e-3t
W [x(1),x(2)](t) =e2t 0
0 e-3t= e-t
Example 2: Direction Field (1 of 9)
• Consider the homogeneous equation x' = Ax below.
• A direction field for this system is given below.
• Substituting x = ert in for x, and rewriting system as
(A – rI) = 0, we obtain
xx
14
11
0
0
14
11
1
1
r
r
x
x
Example 2: Eigenvalues (2 of 9)
• Our solution has the form x = ert, where r and are found
by solving
• Recalling that this is an eigenvalue problem, we determine r
by solving det(A – rI) = 0:
• Thus r1 = 3 and r2 = –1.
0
0
14
11
1
1
r
r
)1)(3(324)1(14
1122
rrrrr
r
r
x x
Example 2: First Eigenvector (3 of 9)
• Eigenvector for r1 = 3: Solve
by row reducing the augmented matrix:
0
0
24
12
0
0
314
131
2
1
2
1
0ξIA r
2
1choosearbitrary,
1
2/12/1
00
02/11
000
02/11
024
02/11
024
012
)1(
2
2)1(
2
21
ξξ cc
Example 2: Second Eigenvector (4 of 9)
• Eigenvector for r2 = -1: Solve
by row reducing the augmented matrix:
0
0
24
12
0
0
114
111
2
1
2
1
0ξIA r
2
1choosearbitrary,
1
2/12/1
00
02/11
000
02/11
024
02/11
024
012
)2(
2
2)2(
2
21
ξξ cc
Example 2: General Solution (5 of 9)
• The corresponding solutions x = ert of x' = Ax are
• The Wronskian of these two solutions is
• Thus x(1) and x(2) are fundamental solutions, and the general
solution of x' = Ax is
tt etet
2
1)(,
2
1)( )2(3)1(
xx
0422
)(, 2
3
3
)2()1(
t
tt
tt
eee
eetW xx
tt ecec
tctct
2
1
2
1
)()()(
2
3
1
)2(
2
)1(
1 xxx
x
Example 2: Phase Plane for x(1) (6 of 9)
• To visualize solution, consider first x = c1x(1):
• Now
• Thus x(1) lies along the straight line x2 = 2x1, which is the line
through origin in direction of first eigenvector (1)
• If solution is trajectory of particle, with position given by
(x1, x2), then it is in Q1 when c1 > 0, and in Q3 when c1 < 0.
• In either case, particle moves away from origin as t increases.
ttt ecxecxecx
xt 3
12
3
11
3
1
2
1)1( 2,2
1)(
x
12
1
2
1
133
12
3
11 22
2, xxc
x
c
xeecxecx ttt
x
Example 2: Phase Plane for x(2) (7 of 9)
• Next, consider x = c2x(2):
• Then x(2) lies along the straight line x2 = –2x1, which is the
line through origin in direction of 2nd eigenvector (2)
• If solution is trajectory of particle, with position given by
(x1, x2), then it is in Q4 when c2 > 0, and in Q2 when c2 < 0.
• In either case, particle moves towards origin as t increases.
ttt ecxecxecx
xt
22212
2
1)2( 2,2
1)(x
x
Example 2:
Phase Plane for General Solution (8 of 9)
• The general solution is x = c1x(1) + c2x
(2):
• As t , c1x(1) is dominant and c2x
(2) becomes negligible.
Thus, for c1 ≠ 0, all solutions asymptotically approach the
line x2 = 2x1 as t .
• Similarly, for c2 ≠ 0, all solutions asymptotically approach
the line x2 = –2x1 as t .
• The origin is a saddle point,
and is unstable. See graph.
tt ecect
2
1
2
1)( 2
3
1x
®¥
®¥
®-¥
Example 2:
Time Plots for General Solution (9 of 9)
• The general solution is x = c1x(1) + c2x
(2):
• As an alternative to phase plane plots, we can graph x1 or x2
as a function of t. A few plots of x1 are given below.
• Note that when c1 = 0, x1(t) = c2e-t 0 as t .
Otherwise, x1(t) = c1e3t + c2e
-t grows unbounded as t .
• Graphs of x2 are similarly obtained.
tt
tt
tt
ecec
ecec
tx
txecect
2
3
1
2
3
1
2
1
2
3
122)(
)(
2
1
2
1)(x
®¥®
®¥
Example 3: Direction Field (1 of 9)
• Consider the homogeneous equation x' = Ax below.
• A direction field for this system is given below.
• Substituting x = ert in for x, and rewriting system as
(A – rI) = 0, we obtain
xx
22
23
-3- r 2
2 -2 - r
æ
èçç
ö
ø÷÷
x1
x2
æ
èç
ö
ø÷ =
0
0
æ
èç
ö
ø÷
x
x
Example 3: Eigenvalues (2 of 9)
• Our solution has the form x = ert, where r and are found
by solving
• Recalling that this is an eigenvalue problem, we determine r
by solving det(A – rI) = 0:
• Thus r1 = –1 and r2 = –4.
)4)(1(452)2)(3(22
23 2
rrrrrr
r
r
-3- r 2
2 -2 - r
æ
èçç
ö
ø÷÷
x1
x2
æ
èç
ö
ø÷ =
0
0
æ
èç
ö
ø÷
x x
Example 3: First Eigenvector (3 of 9)
• Eigenvector for r1 = –1: Solve
by row reducing the augmented matrix:
0
0
12
22
0
0
122
213
2
1
2
1
0ξIA r
2
1choose
2/2
000
02/21
012
02/21
012
022
)1(
2
2)1(ξξ
Example 3: Second Eigenvector (4 of 9)
• Eigenvector for r2 = –4: Solve
by row reducing the augmented matrix:
0
0
22
21
0
0
422
243
2
1
2
1
0ξIA r
1
2choose
2
000
021
022
021
)2(
2
2)2(
ξ
ξ
Example 3: General Solution (5 of 9)
• The corresponding solutions x = ert of x' = Ax are
• The Wronskian of these two solutions is
• Thus x(1) and x(2) are fundamental solutions, and the general
solution of x' = Ax is
tt etet 4)2()1(
1
2)(,
2
1)(
xx
032
2)(, 5
4
4)2()1(
t
tt
tt
eee
eetW xx
tt ecec
tctct
4
21
)2(
2
)1(
1
1
2
2
1
)()()(
xxx
x
Example 3: Phase Plane for x(1) (6 of 9)
• To visualize solution, consider first x = c1x(1):
• Now
• Thus x(1) lies along the straight line x2 = 2½ x1, which is the
line through origin in direction of first eigenvector (1)
• If solution is trajectory of particle, with position given by
(x1, x2), then it is in Q1 when c1 > 0, and in Q3 when c1 < 0.
• In either case, particle moves towards origin as t increases.
ttt ecxecxecx
xt
12111
2
1)1( 2,2
1)(x
12
1
2
1
11211 2
22, xx
c
x
c
xeecxecx ttt
x
Example 3: Phase Plane for x(2) (7 of 9)
• Next, consider x = c2x(2):
• Then x(2) lies along the straight line x2 = –2½ x1, which is the
line through origin in direction of 2nd eigenvector (2)
• If solution is trajectory of particle, with position given by
(x1, x2), then it is in Q4 when c2 > 0, and in Q2 when c2 < 0.
• In either case, particle moves towards origin as t increases.
ttt ecxecxecx
xt 4
22
4
21
4
2
2
1)2( ,21
2)(
x
x
Example 3:
Phase Plane for General Solution (8 of 9)
• The general solution is x = c1x(1) + c2x
(2):
• As t , c1x(1) is dominant and c2x
(2) becomes negligible.
Thus, for c1 ≠ 0, all solutions asymptotically approach
origin along the line x2 = x1 as t .
• Similarly, all solutions are unbounded as t .
• The origin is a node, and is
asymptotically stable.
tt etet 4)2()1(
1
2)(,
2
1)(
xx
®¥
®¥2
®-¥
Example 3:
Time Plots for General Solution (9 of 9)
• The general solution is x = c1x(1) + c2x
(2):
• As an alternative to phase plane plots, we can graph x1 or x2
as a function of t. A few plots of x1 are given below.
• Graphs of x2 are similarly obtained.
tt
tttt
ecec
ecec
tx
txecect
4
21
4
21
2
14
212
2
)(
)(
1
2
2
1)(x
2 x 2 Case:
Real Eigenvalues, Saddle Points and Nodes
• The previous two examples demonstrate the two main cases
for a 2 x 2 real system with real and different eigenvalues:
– Both eigenvalues have opposite signs, in which case origin is a
saddle point and is unstable.
– Both eigenvalues have the same sign, in which case origin is a node,
and is asymptotically stable if the eigenvalues are negative and
unstable if the eigenvalues are positive.
Eigenvalues, Eigenvectors
and Fundamental Solutions
• In general, for an n x n real linear system x' = Ax:
– All eigenvalues are real and different from each other.
– Some eigenvalues occur in complex conjugate pairs.
– Some eigenvalues are repeated.
• If eigenvalues r1,…, rn are real & different, then there are n
corresponding linearly independent eigenvectors (1),…,
(n). The associated solutions of x' = Ax are
• Using Wronskian, it can be shown that these solutions are
linearly independent, and hence form a fundamental set of
solutions. Thus general solution is
trnntr netet )()()1()1( )(,,)( 1 ξxξx
trn
n
tr necec )()1(
11 ξξx
x
x
Hermitian Case: Eigenvalues, Eigenvectors &
Fundamental Solutions
• If A is an n x n Hermitian matrix (real and symmetric), then
all eigenvalues r1,…, rn are real, although some may repeat.
• In any case, there are n corresponding linearly independent
and orthogonal eigenvectors (1),…, (n). The associated
solutions of x' = Ax are
and form a fundamental set of solutions.
trnntr netet )()()1()1( )(,,)( 1 ξxξx
x x
Example 4: Hermitian Matrix (1 of 3)
• Consider the homogeneous equation x' = Ax below.
• The eigenvalues were found previously in Ch 7.3, and were:
r1 = 2, r2 = –1 and r3 = –1.
• Corresponding eigenvectors:
xx
011
101
110
1
1
0
,
1
0
1
,
1
1
1)3()2()1(
ξξξ
Example 4: General Solution (2 of 3)
• The fundamental solutions are
with general solution
ttt eee
1
1
0
,
1
0
1
,
1
1
1)3()2(2)1(
xxx
ttt ececec
1
1
0
1
0
1
1
1
1
32
2
1x
Example 4: General Solution Behavior (3 of 3)
• The general solution is x = c1x(1) + c2x
(2) + c3x(3):
• As t , c1x(1) is dominant and c2x
(2) , c3x(3) become
negligible.
• Thus, for c1 ≠ 0, all solns x become unbounded as t ,
while for c1 = 0, all solns x 0 as t .
• The initial points that cause c1 = 0 are those that lie in plane
determined by (2) and (3). Thus solutions that start in this
plane approach origin as t .
ttt ececec
1
1
0
1
0
1
1
1
1
32
2
1x
®¥
®¥
® ®¥
®¥
x x
Complex Eigenvalues and Fundamental Solns
• If some of the eigenvalues r1,…, rn occur in complex
conjugate pairs, but otherwise are different, then there are
still n corresponding linearly independent solutions
which form a fundamental set of solutions. Some may be
complex-valued, but real-valued solutions may be derived
from them. This situation will be examined in Ch 7.6.
• If the coefficient matrix A is complex, then complex
eigenvalues need not occur in conjugate pairs, but solutions
will still have the above form (if the eigenvalues are
distinct) and these solutions may be complex-valued.
,)(,,)( )()()1()1( 1 trnntr netet ξxξx
Repeated Eigenvalues and Fundamental Solns
• If some of the eigenvalues r1,…, rn are repeated, then there
may not be n corresponding linearly independent solutions of
the form
• In order to obtain a fundamental set of solutions, it may be
necessary to seek additional solutions of another form.
• This situation is analogous to that for an nth order linear
equation with constant coefficients, in which case a repeated
root gave rise solutions of the form
This case of repeated eigenvalues is examined in Section 7.8.
trnntr netet )()()1()1( )(,,)( 1 ξxξx
,,, 2 rtrtrt ettee
Boyce/DiPrima/Meade 11th ed, Ch 7.6:
Complex Eigenvalues
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• We consider again a homogeneous system of n first order
linear equations with constant, real coefficients,
and thus the system can be written as x' = Ax, where
,2211
22221212
12121111
nnnnnn
nn
nn
xaxaxax
xaxaxax
xaxaxax
nnnn
n
n
n aaa
aaa
aaa
tx
tx
tx
t
21
22221
11211
2
1
,
)(
)(
)(
)( Ax
Conjugate Eigenvalues and Eigenvectors
• We know that x = ert is a solution of x' = Ax, provided r is
an eigenvalue and is an eigenvector of A.
• The eigenvalues r1,…, rn are the roots of det(A – rI) = 0,
and the corresponding eigenvectors satisfy (A – rI) = 0.
• If A is real, then the coefficients in the polynomial equation
det(A – rI) = 0 are real, and hence any complex eigenvalues
must occur in conjugate pairs. Thus if r1 = is an
eigenvalue, then so is r2 = .
• The corresponding eigenvectors (1), (2) are conjugates
also.
To see this, recall A and I have real entries, and hence
0ξIA0ξIA0ξIA )2(
2
)1(
1
)1(
1 rrr
x
x
x
l + im
l - im
x x
Conjugate Solutions
• It follows from the previous slide that the solutions
corresponding to these eigenvalues and eigenvectors are
conjugates conjugates as well, since
)1()1()2()2( 22 xξξx trtr
ee
trtree 21 )2()2()1()1( , ξxξx
Example 1: Direction Field (1 of 7)
• Consider the homogeneous equation x' = Ax below.
• A direction field for this system is given below.
• Substituting x = ert in for x, and rewriting system as
(A – rI) = 0, we obtain
xx
2/11
12/1
-1
2- r 1
-1 -1
2- r
æ
è
çççç
ö
ø
÷÷÷÷
x1
x1
æ
èç
ö
ø÷ =
0
0
æ
èç
ö
ø÷
xx
Example 1: Complex Eigenvalues (2 of 7)
• We determine r by solving det(A – rI) = 0. Now
• Thus
• Therefore the eigenvalues are r1 = –1/2 + i and r2 = –1/2 – i.
4
512/1
2/11
12/122
rrr
r
r
ii
r
2
1
2
21
2
)4/5(411 2
Example 1: First Eigenvector (3 of 7)
• Eigenvector for r1 = –1/2 + i: Solve
by row reducing the augmented matrix:
• Thus
0
0
1
1
0
0
1
1
0
0
2/11
12/1
2
1
2
1
1
1
i
i
i
i
r
rr 0ξIA
i
ii
i
i 1choose
000
01
01
01)1(
2
2)1(ξξ
1
0
0
1)1( iξ
Example 1: Second Eigenvector (4 of 7)
• Eigenvector for r1 = –1/2 – i: Solve
by row reducing the augmented matrix:
• Thus
0
0
1
1
0
0
1
1
0
0
2/11
12/1
2
1
2
1
1
1
i
i
i
i
r
rr 0ξIA
i
ii
i
i 1choose
000
01
01
01)2(
2
2)2(ξξ
1
0
0
1)2( iξ
Example 1: General Solution (5 of 7)
• The corresponding solutions x = ert of x' = Ax are
• The Wronskian of these two solutions is
• Thus u(t) and v(t) are real-valued fundamental solutions of
x' = Ax, with general solution x = c1u + c2v.
t
tettet
t
tettet
tt
tt
cos
sincos
1
0sin
0
1)(
sin
cossin
1
0cos
0
1)(
2/2/
2/2/
v
u
0cossin
sincos)(,
2/2/
2/2/
)2()1(
t
tt
tt
etete
tetetW xx
x
Example 1: Phase Plane (6 of 7)
• Given below is the phase plane plot for solutions x, with
• Each solution trajectory approaches origin along a spiral path
as t , since coordinates are products of decaying
exponential and sine or cosine factors.
• The graph of u passes through (1,0),
since u(0) = (1,0). Similarly, the
graph of v passes through (0,1).
• The origin is a spiral point, and
is asymptotically stable.
te
tec
te
tec
x
xt
t
t
t
cos
sin
sin
cos2/
2/
22/
2/
1
2
1x
®¥
Example 1: Time Plots (7 of 7)
• The general solution is x = c1u + c2v:
• As an alternative to phase plane plots, we can graph x1 or x2
as a function of t. A few plots of x1 are given below, each
one a decaying oscillation as t .
tectec
tectec
tx
txtt
tt
cossin
sincos
)(
)(2/
2
2/
1
2/
2
2/
1
2
1x
®¥
General Solution
• To summarize, suppose r1 = , r2 = , and that
r3,…, rn are all real and distinct eigenvalues of A. Let the
corresponding eigenvectors be
• Then the general solution of x' = Ax is
where
trn
n
tr necectctc )()3(
3213)()( ξξvux
)()4()3()2()1( ,,,,, nii ξξξbaξbaξ
ttetttet tt cossin)(,sincos)( bavbau
l + im l - im
Real-Valued Solutions
• Thus for complex conjugate eigenvalues r1 and r2 , the corresponding solutions x(1) and x(2) are conjugates also.
• To obtain real-valued solutions, use real and imaginary parts of either x(1) or x(2). To see this, let (1) = a + i b. Then
where
are real valued solutions of x' = Ax, and can be shown to be linearly independent.
)()(
cossinsincos
sincos)1()1(
tit
ttiette
titeie
tt
tti
vu
baba
baξx
,cossin)(,sincos)( ttetttet tt bavbau
x
Spiral Points, Centers,
Eigenvalues, and Trajectories
• In previous example, general solution was
• The origin was a spiral point, and was asymptotically stable.
• If real part of complex eigenvalues is positive, then
trajectories spiral away, unbounded, from origin, and hence
origin would be an unstable spiral point.
• If real part of complex eigenvalues is zero, then trajectories
circle origin, neither approaching nor departing. Then origin
is called a center and is stable, but not asymptotically stable.
Trajectories periodic in time.
• The direction of trajectory motion depends on entries in A.
te
tec
te
tec
x
xt
t
t
t
cos
sin
sin
cos2/
2/
22/
2/
1
2
1x
Example 2:
Second Order System with Parameter (1 of 2)
• The system x' = Ax below contains a parameter .
• Substituting x = ert in for x and rewriting system as
(A – rI) = 0, we obtain
• Next, solve for r in terms of :
xx
02
2
a - r 2
-2 -r
æ
èç
ö
ø÷
x1
x2
æ
èç
ö
ø÷ =
0
0
æ
èç
ö
ø÷
2
1644)(
2
2 22
rrrrr
r
r
xx
a
a
Example 2:
Eigenvalue Analysis (2 of 2)
• The eigenvalues are given by the quadratic formula above.
• For < –4, both eigenvalues are real and negative, and hence origin is asymptotically stable node.
• For > 4, both eigenvalues are real and positive, and hence the origin is an unstable node.
• For –4 < < 0, eigenvalues are complex with a negative real part, and hence origin is asymptotically stable spiral point.
• For 0 < < 4, eigenvalues are complex with a positive real part, and the origin is an unstable spiral point.
• For = 0, eigenvalues are purely imaginary, origin is a center. Trajectories closed curves about origin & periodic.
• For = ± 4, eigenvalues real & equal, origin is a node (Ch 7.8)
2
162
r
a
a
a
a
a
a
Second Order Solution Behavior and
Eigenvalues: Three Main Cases
• For second order systems, the three main cases are:
– Eigenvalues are real and have opposite signs; x = 0 is a saddle point.
– Eigenvalues are real, distinct and have same sign; x = 0 is a node.
– Eigenvalues are complex with nonzero real part; x = 0 a spiral point.
• Other possibilities exist and occur as transitions between two
of the cases listed above:
– A zero eigenvalue occurs during transition between saddle point and
node. Real and equal eigenvalues occur during transition between
nodes and spiral points. Purely imaginary eigenvalues occur during a
transition between asymptotically stable and unstable spiral points.
a
acbbr
2
42
Example 3: Multiple Spring-Mass System (1 of 6)
• The equations for the system of two masses and three
springs discussed in Section 7.1, assuming no external
forces, can be expressed as:
• Given , , the
equations become
' and,',, where
)(' and)('or
)( and)(
24132211
23212422212131
232122
2
2
2221212
1
2
1
xyxyxyxy
ykkykymykykkym
xkkxkdt
xdmxkxkk
dt
xdm
4/15 and,3,1,4/9,2 32121 kkkmm
2142134231 33/4' and,2/32',',' yyyyyyyyyy
Example 3: Multiple Spring-Mass System (2 of 6)
• Writing the system of equations in matrix form:
• Assuming a solution of the form y = ert , where r must
be an eigenvalue of the matrix A and is the
corresponding eigenvector, the characteristic polynomial of
A is
yielding the eigenvalues:
)4)(1(45 2224 rrrr
iriririr 2 and,2,, 4321
2142134231 33/4' and,2/32',',' yyyyyyyyyy
xx
Example 3: Multiple Spring-Mass System (3 of 6)
• For the eigenvalues the correspond-
ing eigenvectors are
• The products yield the complex-valued solutions:
Ayyy'
0033/4
002/32
1000
0100
iriririr 2 and,2,, 4321
i
i
i
i
i
i
i
i
8
6
4
3
and,
8
6
4
3
,
2
3
2
3
,
2
3
2
3
)4()3()2()1(
itit ee 2)3()1( and ξξ
)()(
2cos8
2cos6
2sin4
2sin3
2sin8
2sin6
2cos4
2cos3
)2sin2(cos
8
6
4
3
)()(
cos2
cos3
sin2
sin3
sin2
sin3
cos2
cos3
)sin(cos
2
3
2
3
)2()2(2)3(
)1()1()1(
tit
t
t
t
t
i
t
t
t
t
tit
i
ie
tit
t
t
t
t
i
t
t
t
t
tit
i
ie
it
it
vu
vu
Example 3: Multiple Spring-Mass System (4 of 6)
• After validating that are linearly
independent, the general solution of the system of equations can be
written as
• where are arbitrary constants.
• Each solution will be periodic with period 2π, so each trajectory is a
closed curve. The first two terms of the solution describe motions with
frequency 1 and period 2π while the second two terms describe
motions with frequency 2 and period π. The motions of the two masses
will be different relative to one another for solutions involving only the
first two terms or the second two terms.
)(),(,)(),( )2()2()1()1( tttt vuvu
y = c1
3cost
2cos t
-3sin t
-2sin t
æ
è
çççç
ö
ø
÷÷÷÷
+ c2
3sin t
2sin t
3cos t
2cost
æ
è
çççç
ö
ø
÷÷÷÷
+ c3
3cos2t
-4 cos2t
-6sin2t
8sin2t
æ
è
çççç
ö
ø
÷÷÷÷
+ c4
3sin2t
-4sin2t
6cos2t
-8cos2t
æ
è
çççç
ö
ø
÷÷÷÷
4321 ,,, cccc
2142134231 33/4' and,2/32',',' yyyyyyyyyy
Example 3: Multiple Spring-Mass System (5 of 6)
• To obtain the fundamental mode of vibration with frequency 1
• To obtain the fundamental mode of vibration with frequency 2
• Plots of and parametric plots (y, y’) are shown for a
selected solution with frequency 1
)0(2)0(3 and)0(2)0(3 when occurs 0 341243 yyyycc
)0(4)0(3 and)0(4)0(3 when occurs 0 341221 yyyycc
2y
',' and masses theofmotion therepresent and 141321yyyyyy
21 and yy
Example 3: Multiple Spring-Mass System (6 of 6)
• Plots of and parametric plots (y, y’) are shown for a selected
solution with frequency 2
',' and masses theofmotion therepresent and 141321yyyyyy
21 and yy
Boyce/DiPrima/Meade 11th ed, Ch 7.7: Fundamental Matrices
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• Suppose that x(1)(t),…, x(n)(t) form a fundamental set of
solutions for x' = P(t)x on .
• The matrix
whose columns are x(1)(t),…, x(n)(t), is a fundamental matrix
for the system x' = P(t)x. This matrix is nonsingular since its
columns are linearly independent, and hence det ≠ 0.
• Note also that since x(1)(t),…, x(n)(t) are solutions of x' = P(t)x,
satisfies the matrix differential equation ' = P(t) .
,
)()(
)()(
)()()1(
)(
1
)1(
1
txtx
txtx
tn
nn
n
Ψ
a < t < b
Y
Y Y Y
Example 1:
• Consider the homogeneous equation x' = Ax below.
• In Example 2 of Chapter 7.5, we found the following
fundamental solutions for this system:
• Thus a fundamental matrix for this system is
xx
14
11
tt etet
2
1)(,
2
1)( )2(3)1(
xx
tt
tt
ee
eet
22)(
3
3
Ψ
Fundamental Matrices and General Solution
• The general solution of x' = P(t)x
can be expressed x = (t)c, where c is a constant vector with
components c1,…, cn:
)()1(
1 )( n
nctc xxx
n
n
nn
n
c
c
txtx
txtx
t
1
)()1(
)(
1
)1(
1
)()(
)()(
)( cΨx
Y
Fundamental Matrix & Initial Value Problem
• Consider an initial value problem
x' = P(t)x, x(t0) = x0
where and x0 is a given initial vector.
• Now the solution has the form x = (t)c, hence we choose c
so as to satisfy x(t0) = x0.
• Recalling (t0) is nonsingular, it follows that
• Thus our solution x = (t)c can be expressed as
0
0
10
0 )()( xΨcxcΨ tt
0
0
1 )()( xΨΨx tt
Y
a < t0 < b
Y
Y
Recall: Theorem 7.4.4
• Let
• Let x(1),…, x(n) be solutions of x' = P(t)x on I:
that satisfy the initial conditions
Then x(1),…, x(n) are fundamental solutions of x' = P(t)x.
1
0
0
0
,,
0
0
1
0
,
0
0
0
1
)()2()1(
neee
0
)(
0
)()1(
0
)1( ,)(,,)( ttt nnexex
a < t < b
Fundamental Matrix & Theorem 7.4.4
• Suppose x(1)(t),…, x(n)(t) form the fundamental solutions given
by Thm 7.4.4. Denote the corresponding fundamental matrix
by (t). Then columns of (t) are x(1)(t),…, x(n)(t), and
hence
• Thus –1(t0) = I, and the hence general solution to the
corresponding initial value problem is
• It follows that for any fundamental matrix (t),
I
100
010
001
)( 0
t
00
0
1 )()()( xΦxΦΦx ttt
)()()()()()( 0
100
0
1 tttttt ΨΨΦxΦxΨΨx
F F
F
F
The Fundamental Matrix .
and Varying Initial Conditions
• Thus when using the fundamental matrix (t), the general
solution to an IVP is
• This representation is useful if same system is to be solved for
many different initial conditions, such as a physical system
that can be started from many different initial states.
• Also, once (t) has been determined, the solution to each set
of initial conditions can be found by matrix multiplication, as
indicated by the equation above.
• Thus (t) represents a linear transformation of the initial
conditions x0 into the solution x(t) at time t.
00
0
1 )()()( xΦxΦΦx ttt
F
F
F
F
Example 2: Find (t) for 2 x 2 System (1 of 5)
• Find (t) such that (0) = I for the system below.
• Solution: First, we must obtain x(1)(t) and x(2)(t) such that
• We know from previous results that the general solution is
• Every solution can be expressed in terms of the general
solution, and we use this fact to find x(1)(t) and x(2)(t).
xx
14
11
tt ecec
2
1
2
12
3
1x
1
0)0(,
0
1)0( )2()1(
xx
F
F F
Example 2: Use General Solution (2 of 5)
• Thus, to find x(1)(t), express it terms of the general solution
and then find the coefficients c1 and c2.
• To do so, use the initial conditions to obtain
or equivalently,
tt ecect
2
1
2
1)( 2
3
1
)1(x
0
1
2
1
2
1)0( 21
)1( ccx
0
1
22
11
2
1
c
c
Example 2: Solve for x(1)(t) (3 of 5)
• To find x(1)(t), we therefore solve
by row reducing the augmented matrix:
• Thus
2/1
2/1
2/110
2/101
2/110
111
240
111
022
111
2
1
c
c
tt
tttt
ee
eeeet
3
33)1(
2
1
2
1
2
1
2
1
2
1
2
1)(x
0
1
22
11
2
1
c
c
Example 2: Solve for x(2)(t) (4 of 5)
• To find x(2)(t), we similarly solve
by row reducing the augmented matrix:
• Thus
4/1
4/1
4/110
4/101
4/110
011
140
011
122
011
2
1
c
c
tt
tt
tt
ee
eeeet
2
1
2
14
1
4
1
2
1
4
1
2
1
4
1)(
3
3
3)2(x
1
0
22
11
2
1
c
c
Example 2: Obtain (t) (5 of 5)
• The columns of (t) are given by x(1)(t) and x(2)(t), and
thus from the previous slide we have
• Note (t) is more complicated than (t) found in Ex 1.
However, it is now much easier to determine the solution to
any set of initial conditions.
tttt
tttt
eeee
eeeet
2
1
2
14
1
4
1
2
1
2
1
)(33
33
Φ
tt
tt
ee
eet
22)(
3
3
Ψ
F
F Y
F
Matrix Exponential Functions
• Consider the following two cases:
– The solution to x' = ax, x(0) = x0, is x = x0eat, where e0 = 1.
– The solution to x' = Ax, x(0) = x0, is x = (t)x0, where (0) = I.
• Comparing the form and solution for both of these cases, we
might expect (t) to have an exponential character.
• Indeed, it can be shown that (t) = eAt, where
is a well defined matrix function that has all the usual
properties of an exponential function. See text for details.
• Thus the solution to x' = Ax, x(0) = x0, is x = eAtx0.
10 !! n
nn
n
nnt
n
t
n
te
AI
AA
F F
F
F
Coupled Systems of Equations
• Recall that our constant coefficient homogeneous system
written as x' = Ax with
is a system of coupled equations that must be solved
simultaneously to find all the unknown variables.
,2211
12121111
nnnnnn
nn
xaxaxax
xaxaxax
,,
)(
)(
)(
1
1111
nnn
n
n aa
aa
tx
tx
t
Ax
Uncoupled Systems & Diagonal Matrices
• In contrast, if each equation had only one variable, solved for
independently of other equations, then task would be easier.
• In this case our system would have the form
or x' = Dx, where D is a diagonal matrix:
,00
00
00
21
21112
21111
nnnn
n
n
xdxxx
xxdxx
xxxdx
nn
nd
d
d
tx
tx
t
00
00
00
,
)(
)(
)(22
11
1
Dx
Uncoupling: Transform Matrix T
• In order to explore transforming our given system x' = Ax of
coupled equations into an uncoupled system x' = Dx, where D is
a diagonal matrix, we will use the eigenvectors of A.
• Suppose A is n x n with n linearly independent eigenvectors
, and corresponding eigenvalues .
• Define n x n matrices T and D using the eigenvalues &
eigenvectors of A:
• Note that T is nonsingular, and hence T-1 exists.
n
n
nn
n
00
00
00
,2
1
)()1(
)(
1
)1(
1
DT
x (1),...,x (n) l1,...,ln
Uncoupling: T-1AT = D
• Recall here the definitions of A, T and D:
• Then the columns of AT are A (1),…, A (n), and hence
• It follows that T–1AT = D.
n
n
nn
n
nnn
n
aa
aa
00
00
00
,,2
1
)()1(
)(
1
)1(
1
1
111
DTA
TDAT
)()1(
1
)(
1
)1(
11
n
nnn
n
n
x x
Similarity Transformations
• Thus, if the eigenvalues and eigenvectors of A are known,
then A can be transformed into a diagonal matrix D, with
T–1AT = D
• This process is known as a similarity transformation, and A
is said to be similar to D. Alternatively, we could say that A
is diagonalizable.
n
n
nn
n
nnn
n
aa
aa
00
00
00
,,2
1
)()1(
)(
1
)1(
1
1
111
DTA
Similarity Transformations: Hermitian Case
• Recall: Our similarity transformation of A has the form
T–1AT = D
where D is diagonal and columns of T are eigenvectors of A.
• If A is Hermitian, then A has n linearly independent
orthogonal eigenvectors , normalized so that
for i = 1,…, n, and for i ≠ k.
• With this selection of eigenvectors, it can be shown that
T–1 = T*. In this case we can write our similarity transform
as
T*AT = D
x (1),...,x (n)
(x (i ),x (i )) = 1 (x (i ),x (k )) = 0
Nondiagonalizable A
• Finally, if A is n x n with fewer than n linearly independent
eigenvectors, then there is no matrix T such that T–1AT = D.
• In this case, A is not similar to a diagonal matrix and A is not
diagonlizable.
n
n
nn
n
nnn
n
aa
aa
00
00
00
,,2
1
)()1(
)(
1
)1(
1
1
111
DTA
Example 3:
Find Transformation Matrix T (1 of 2)
• For the matrix A below, find the similarity transformation
matrix T and show that A can be diagonalized.
• We already know that the eigenvalues are 1 = 3, 2 = –1
with corresponding eigenvectors
• Thus
14
11A
2
1)(,
2
1)( )2()1( tt ξξ
10
03,
22
11DT
l l
Example 3: Similarity Transformation (2 of 2)
• To find T–1, augment the identity to T and row reduce:
• Then
• Thus A is similar to D, and hence A is diagonalizable.
4/12/1
4/12/1
4/12/110
4/12/101
4/12/110
0111
1240
0111
1022
0111
1T
D
ATT
10
03
26
13
4/12/1
4/12/1
22
11
14
11
4/12/1
4/12/11
Fundamental Matrices for Similar Systems (1 of 3)
• Recall our original system of differential equations x' = Ax.
• If A is n x n with n linearly independent eigenvectors, then A
is diagonalizable. The eigenvectors form the columns of the
nonsingular transform matrix T, and the eigenvalues are the
corresponding nonzero entries in the diagonal matrix D.
• Suppose x satisfies x' = Ax, let y be the n x 1 vector such that
x = Ty. That is, let y be defined by y = T–1x.
• Since x' = Ax and T is a constant matrix, we have Ty' = ATy,
and hence y' = T–1ATy = Dy.
• Therefore y satisfies y' = Dy, the system similar to x' = Ax.
• Both of these systems have fundamental matrices, which we
examine next.
Fundamental Matrix for Diagonal System (2 of 3)
• A fundamental matrix for y' = Dy is given by Q(t) = eDt.
• Recalling the definition of eDt, we have
t
t
n
n
n
n
n
n
n
n
n
n
n
nn
ne
e
n
t
n
t
n
t
n
tt
00
00
00
!00
00
00!
!00
00
00
!)(
1
0
0
1
0
1
0
D
Q
Fundamental Matrix for Original System (3 of 3)
• To obtain a fundamental matrix (t) for x' = Ax, recall that
the columns of (t) consist of fundamental solutions x
satisfying x' = Ax. We also know x = Ty, and hence it follows
that
• The columns of (t) given the expected fundamental
solutions of x' = Ax.
tn
n
t
n
tnt
t
t
n
nn
n
n
n
n ee
ee
e
e
)()1(
)(
1
)1(
1
)()1(
)(
1
)1(
1
1
11
00
00
00
TQΨ
YY
Y
Example 4:
Fundamental Matrices for Similar Systems
• We now use the analysis and results of the last few slides.
• Applying the transformation x = Ty to x' = Ax below, this
system becomes y' = T–1ATy = Dy:
• A fundamental matrix for y' = Dy is given by Q(t) = eDt:
• Thus a fundamental matrix (t) for x' = Ax is
yyxx
10
03
14
11
t
t
e
et
0
0)(
3
Q
tt
tt
t
t
ee
ee
e
et
220
0
22
11)(
3
33
TQΨ
Y
Boyce/DiPrima/Meade 11th ed, Ch 7.8:
Repeated Eigenvalues
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• We consider again a homogeneous system of n first order
linear equations with constant real coefficients x' = Ax.
• If the eigenvalues r1,…, rn of A are real and different, then
there are n linearly independent eigenvectors ,
and n linearly independent solutions of the form
• If some of the eigenvalues r1,…, rn are repeated, then there
may not be n corresponding linearly independent solutions of
the above form.
• In this case, we will seek additional solutions that are products
of polynomials and exponential functions.
trnntr netet )()()1()1( )(,,)( 1 ξxξx
x (1),...,x (n)
Example 1: Eigenvalues (1 of 2)
• We need to find the eigenvectors for the matrix:
• The eigenvalues r and eigenvectors satisfy the equation
(A – rI ) =0 or
• To determine r, solve det(A – rI) = 0:
• Thus r1 = 2 and r2 = 2.
22 )2(441)3)(1(31
11
rrrrr
r
r
0
0
31
11
1
1
r
r
A =1 -1
1 3
æ
èç
ö
ø÷x
xx
Example 1: Eigenvectors (2 of 2)
• To find the eigenvectors, we solve
by row reducing the augmented matrix:
• Thus there is only one linearly independent eigenvector for
the repeated eigenvalue r = 2.
0
0
11
11
0
0
231
121
2
1
2
1
0ξIA r
1
1choose
00
011
000
011
011
011
011
011
)1(
2
2)1(
2
21
ξξ
Example 2: Direction Field (1 of 10)
• Consider the homogeneous equation x' = Ax below.
• A direction field for this system is given below.
• Substituting x = ert in for x, where r is A’s eigenvalue and
is its corresponding eigenvector,
the previous example showed the
existence of only one eigenvalue,
r = 2, with one eigenvector:
xx
31
11
1
1ξ
xx
Example 2: First Solution; and
Second Solution, First Attempt (2 of 10)
• The corresponding solution x = ert of x' = Ax is
• Since there is no second solution of the form x = ert, we
need to try a different form. Based on methods for second
order linear equations in Ch 3.5, we first try x = te2t.
• Substituting x = te2t into x' = Ax, we obtain
or
tet 2)1(
1
1)(
x
ttt tetee 222 2 Aξξξ
02 222 ttt teete Aξξξ
x
x
x
x
Example 2:
Second Solution, Second Attempt (3 of 10)
• From the previous slide, we have
• In order for this equation to be satisfied for all t, it is
necessary for the coefficients of te2t and e2t to both be zero.
• From the e2t term, we see that = 0, and hence there is no
nonzero solution of the form x = te2t.
• Since te2t and e2t appear in the above equation, we next
consider a solution of the form
02 222 ttt teete Aξξξ
tt ete 22 ηξx
xx
Example 2: Second Solution and its
Defining Matrix Equations (4 of 10)
• Substituting x = te2t + e2t into x' = Ax, we obtain
or
• Equating coefficients yields A = 2 and A = + 2 ,
or
• The first equation is satisfied if is an eigenvector of A
corresponding to the eigenvalue r = 2. Thus
ttttt eteetee 22222 22 ηξAηξξ
tttt eteete 2222 22 AηAξηξξ
ξηIA0ξIA 2and2
1
1ξ
x h
x x xh h
x
Example 2: Solving for Second Solution (5 of 10)
• Recall that
• Thus to solve (A – 2I) = for , we row reduce the
corresponding augmented matrix:
1
1
1
0
1
1000
111
111
111
111
111
1
1
12
kηη
1
1,
31
11ξA
h x h
Example 2: Second Solution (6 of 10)
• Our second solution x = te2t + e2t is now
• Recalling that the first solution was
we see that our second solution is simply
since the last term of third term of x is a multiple of x(1).
ttt ekete 222
1
1
1
0
1
1
x
,1
1)( 2)1( tet
x
,1
0
1
1)( 22)2( tt etet
x
hx
Example 2: General Solution (7 of 10)
• The two solutions of x' = Ax are
• The Wronskian of these two solutions is
• Thus x(1) and x(2) are fundamental solutions, and the general
solution of x' = Ax is
0)(, 4
222
22
)2()1(
t
ttt
tt
eetee
teetW xx
ttt etecec
tctct
22
2
2
1
)2(
2
)1(
1
1
0
1
1
1
1
)()()( xxx
ttt etetet 22)2(2)1(
1
0
1
1)(,
1
1)(
xx
Example 2: Phase Plane (8 of 10)
• The general solution is
• Thus x is unbounded as t , and x 0 as t .
• Further, it can be shown that as t , x 0 asymptotic
to the line x2 = –x1 determined by the first eigenvector.
• Similarly, as t , x is asymptotic
to a line parallel to x2 = –x1.
ttt etecect 22
2
2
11
0
1
1
1
1)(x
®¥®®-¥
®-¥®
®¥
Example 2: Phase Plane (9 of 10)
• The origin is an improper node, and is unstable. See graph.
• The pattern of trajectories is typical for two repeated
eigenvalues with only one eigenvector.
• If the eigenvalues are negative, then the trajectories are
similar but are traversed in the inward direction. In this case
the origin is an asymptotically stable improper node.
Example 2:
Time Plots for General Solution (10 of 10)
• Time plots for x1(t) are given below, where we note that the
general solution x can be written as follows.
tt
tt
ttt
tececc
tecec
tx
tx
etecect
2
2
2
21
2
2
2
1
2
1
22
2
2
1
)()(
)(
1
0
1
1
1
1)(x
General Case for Double Eigenvalues
• Suppose the system x' = Ax has a double eigenvalue r = and a single corresponding eigenvector .
• The first solution is
,
where satisfies (A – I) = 0.
• As in Example 1, the second solution has the form
where is as above and satisfies .
• Even though det(A – I) = 0, it can be shown that
can always be solved for .
• The vector is called a generalized eigenvector.
tt ete ηξx )2(
rx
x(1)(t) = xert
x r x
x h
h
r
h(A - rI)h = x
(A - rI)h = x
Example 2 Extension:
Fundamental Matrix (1 of 2)
• Recall that a fundamental matrix (t) for x' = Ax has
linearly independent solution for its columns.
• In Example 1, our system x' = Ax was
and the two solutions we found were
• Thus the corresponding fundamental matrix is
11
1)( 2
222
22
t
te
etee
teet t
ttt
tt
Ψ
x(1)(t) =
1
-1
æ
èç
ö
ø÷e2t , x(2)(t) =
1
-1
æ
èç
ö
ø÷te2t +
0
-1
æ
èç
ö
ø÷e2t
xx
31
11
Y
Example 2 Extension:
Fundamental Matrix (2 of 2)
• The fundamental matrix (t) that satisfies (0) = I can be
found using , where
where –1(0) is found as follows:
• Thus
,11
01)0(,
11
01)0( 1
ΨΨ
1110
0101
1110
0101
1011
0101
1
1
11
01
11
1)( 22
tt
tte
t
tet tt
Φ
FF(t) = Y(t)Y-1(0)
F
Y
Jordan Forms
• If A is n x n with n linearly independent eigenvectors, then A
can be diagonalized using a similarity transform T–1AT = D. The transform matrix T consisted of eigenvectors of A, and
the diagonal entries of D consisted of the eigenvalues of A.
• In the case of repeated eigenvalues and fewer than n linearly
independent eigenvectors, A can be transformed into a nearly
diagonal matrix J, called the Jordan form of A, with
T–1AT = J.
Example 2 Extension:
Transform Matrix (1 of 2)
• In Example 2, our system x' = Ax was
with eigenvalues r1 = 2 and r2 = 2 and eigenvectors
• Choosing k = 0, the transform matrix T formed from the
two eigenvectors and is
xx
31
11
1
1
1
0,
1
1kηξ
11
01T
x h
Example 2 Extension: Jordan Form (2 of 2)
• The Jordan form J of A is defined by T–1AT = J.
• Now
and hence
• Note that the eigenvalues of A, r1 = 2 and r2 = 2, are on the
main diagonal of J, and that there is a 1 directly above the
second eigenvalue. This pattern is typical of Jordan forms.
11
01,
11
011
TT
20
12
11
01
31
11
11
011ATTJ
Boyce/DiPrima/Meade 11th ed, Ch 7.9: Nonhomogeneous
Linear Systems
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• The general theory of a nonhomogeneous system of equations
parallels that of a single nth order linear equation.
• This system can be written as x' = P(t)x + g(t), where
)()()()(
)()()()(
)()()()(
2211
222221212
112121111
tgxtpxtpxtpx
tgxtpxtpxtpx
tgxtpxtpxtpx
nnnnnnn
nn
nn
)()()(
)()()(
)()()(
)(,
)(
)(
)(
)(,
)(
)(
)(
)(
21
22221
11211
2
1
2
1
tptptp
tptptp
tptptp
t
tg
tg
tg
t
tx
tx
tx
t
nnnn
n
n
nn
Pgx
General Solution
• The general solution of x' = P(t)x + g(t) on I:
has the form
where
is the general solution of the homogeneous system x' = P(t)x
and v(t) is a particular solution of the nonhomogeneous
system x' = P(t)x + g(t).
)()()()( )()2(
2
)1(
1 ttctctc n
n vxxxx
)()()( )()2(
2
)1(
1 tctctc n
nxxx
a < t < b
Diagonalization
• Suppose x' = Ax + g(t), where A is an n x n diagonalizable
constant matrix.
• Let T be the nonsingular transform matrix whose columns are
the eigenvectors of A, and D the diagonal matrix whose
diagonal entries are the corresponding eigenvalues of A.
• Suppose x satisfies x' = Ax, let y be defined by x = Ty.
• Substituting x = Ty into x' = Ax, we obtain
Ty' = ATy + g(t).
or y' = T–1ATy + T–1g(t)
or y' = Dy + h(t), where h(t) = T–1g(t).
• Note that if we can solve diagonal system y' = Dy + h(t) for y,
then x = Ty is a solution to the original system.
Solving Diagonal System
• Now y' = Dy + h(t) is a diagonal system of the form
where r1,…, rn are the eigenvalues of A.
• Thus y' = Dy + h(t) is an uncoupled system of n linear first
order equations in the unknowns yk(t), which can be isolated
and solved separately, using methods of Section 2.1:
nnnnnnn
n
n
h
h
h
y
y
y
r
r
r
y
y
y
thyryyy
thyyryy
thyyyry
2
1
2
1
2
1
2
1
121
22212
12111
00
00
00
)(00
)(00
)(00
nkthyry kkk ,,1),(1
nkecdssheeyt
t
tr
kk
srtr
kkkk ,,1,)(
0
Solving Original System
• The solution y to y' = Dy + h(t) has components
• For this solution vector y, the solution to the original system
x' = Ax + g(t) is then x = Ty.
• Recall that T is the nonsingular transform matrix whose
columns are the eigenvectors of A.
• Thus, when multiplied by T, the second term on right side of
yk produces general solution of homogeneous equation, while
the integral term of yk produces a particular solution of
nonhomogeneous system.
nkecdssheeyt
t
tr
kk
srtr
kkkk ,,1,)(
0
Example 1: General Solution of Homogeneous
Case (1 of 5)
• Consider the nonhomogeneous system x' = Ax + g below.
• Note: A is a Hermitian matrix, since it is real and symmetric.
• The eigenvalues of A are r1 = -3 and r2 = -1, with
corresponding eigenvectors
• The general solution of the homogeneous system is then
¢x =-2 1
1 -2
æ
èç
ö
ø÷x+
2e-t
3t
æ
èç
ö
ø÷ = Ax + g(t)
1
1,
1
1)2()1(
ξξ
x(t) = c1
1
-1
æ
èç
ö
ø÷e-3t + c2
1
1
æ
èç
ö
ø÷e-t
Example 1: Transformation Matrix (2 of 5)
• Consider next the transformation matrix T of eigenvectors.
Using a Section 7.7 comment, and A Hermitian, we have
T–1 = T* = TT, provided we normalize (1)and (2) so that ( (1), (1)) = 1 and ( (2), (2)) = 1. Thus normalize as follows:
• Then for this choice of eigenvectors,
1
1
2
1
1
1
)1)(1()1)(1(
1
,1
1
2
1
1
1
)1)(1()1)(1(
1
)2(
)1(
ξ
ξ
11
11
2
1,
11
11
2
1 1TT
x xx x x x
Example 1:
Diagonal System and its Solution (3 of 5)
• Under the transformation x = Ty, we obtain the diagonal
system y' = Dy + T–1g(t):
• Then, using methods of Section 2.1,
te
te
y
y
t
e
y
y
y
y
t
t
t
32
32
2
13
3
2
11
11
2
1
10
03
2
1
2
1
2
1
ttt
ttt
ectteyteyy
ect
eyteyy
2222
3
1111
12
32
2
32
9
1
32
3
2
2
2
323
Example 1:
Transform Back to Original System (4 of 5)
• We next use the transformation x = Ty to obtain the solution
to the original system x' = Ax + g(t):
2,
2,
3
52
2
1
3
4
2
1
12
36
1
22
1
11
11
11
11
2
1
22
11
2
3
1
2
3
1
2
3
1
2
1
2
1
ck
ck
tetekek
tetekek
ektte
ekt
e
y
y
x
x
ttt
ttt
tt
tt
Example 1:
Solution of Original System (5 of 5)
• Simplifying further, the solution x can be written as
• Note that the first two terms on right side form the general
solution to homogeneous system, while the remaining terms
are a particular solution to nonhomogeneous system.
5
4
3
1
2
1
1
1
1
1
2
1
1
1
1
1
3
52
2
1
3
4
2
1
2
3
1
2
3
1
2
3
1
2
1
tteeekek
tetekek
tetekek
x
x
tttt
ttt
ttt
Nondiagonal Case
• If A cannot be diagonalized, (repeated eigenvalues and a
shortage of eigenvectors), then it can be transformed to its
Jordan form J, which is nearly diagonal.
• In this case the differential equations are not totally
uncoupled, because some rows of J have two nonzero
entries: an eigenvalue in diagonal position, and a 1 in
adjacent position to the right of diagonal position.
• However, the equations for y1,…, yn can still be solved
consecutively, starting with yn. Then the solution x to
original system can be found using x = Ty.
Undetermined Coefficients
• A second way of solving x' = P(t)x + g(t) is the method of
undetermined coefficients. Assume P is a constant matrix,
and that the components of g are polynomial, exponential or
sinusoidal functions, or sums or products of these.
• The procedure for choosing the form of solution is usually
directly analogous to that given in Section 3.6.
• The main difference arises when g(t) has the form ue t,
where is a simple eigenvalue of P. In this case, g(t)
matches solution form of homogeneous system x' = P(t)x,
and as a result, it is necessary to take nonhomogeneous
solution to be of the form ate t + be t. This form differs
from the Section 3.6 analog, ate t.
l
l
l l
l
Example 2: Undetermined Coefficients (1 of 5)
• Consider again the nonhomogeneous system x' = Ax + g:
• Assume a particular solution of the form
where the vector coefficents a, b, c, d are to be determined.
• Since r = -1 is an eigenvalue of A, it is necessary to include
both ate-t and be-t, as mentioned on the previous slide.
¢x =-2 1
1 -2
æ
èç
ö
ø÷x+
2e-t
3t
æ
èç
ö
ø÷ =
-2 1
1 -2
æ
èç
ö
ø÷x+
2
0
æ
èç
ö
ø÷e-t +
0
3
æ
èç
ö
ø÷t
dcbav tetet tt)(
Example 2:
Matrix Equations for Coefficients (2 of 5)
• Substituting
in for x in our nonhomogeneous system x' = Ax + g,
we obtain
• Equating coefficients, we conclude that
¢x =-2 1
1 -2
æ
èç
ö
ø÷x +
2
0
æ
èç
ö
ø÷e-t +
0
3
æ
èç
ö
ø÷t,
dcbav tetet tt)(
Aa = -a, Ab = a - b -2
0
æ
èç
ö
ø÷, Ac = -
0
3
æ
èç
ö
ø÷, Ad = c
-ate-t + a - b( )e-t + c = Aate-t +Abe-t +Act +Ad +2
0
æ
èç
ö
ø÷e-t +
0
3
æ
èç
ö
ø÷t
Example 2:
Solving Matrix Equation for (a) (3 of 5)
• Our matrix equations for the coefficients are:
• From the first equation, we see that a is an eigenvector of A
corresponding to eigenvalue r = -1, and hence has the form
• We will see on the next slide that = 1, and hence
cAdAcbaAbaAa
,
3
0,
0
2,
a
1
1a
a
Example 2:
Solving Matrix Equation for (b) (4 of 5)
• Our matrix equations for the coefficients are:
• Substituting a = into the second equation,
• Thus = 1, and solving for b, we obtain
cAdAcbaAbaAa
,
3
0,
0
2,
100
112
11
11
2
10
01
21
122
2
1
2
1
2
1
2
1
b
b
b
b
b
b
b
bAb
1
00 choose
1
0
1
1bb kk
(a ,a)T
a
Example 2: Particular Solution (5 of 5)
• Our matrix equations for the coefficients are:
• Solving third equation for c, and then fourth equation for d,
it is straightforward to obtain cT = (1, 2), dT = (–4/3, –5/3).
• Thus our particular solution of x' = Ax + g is
• Comparing this to the result obtained in Example 1, we see
that both particular solutions would be the same if we had
chosen k = ½ for b on previous slide, instead of k = 0.
cAdAcbaAbaAa
,
3
0,
0
2,
v(t) =1
1
æ
èç
ö
ø÷te-t -
0
1
æ
èç
ö
ø÷e- t +
1
2
æ
èç
ö
ø÷t -
1
3
4
5
æ
èç
ö
ø÷
Variation of Parameters: Preliminaries
• A more general way of solving x' = P(t)x + g(t) is the
method of variation of parameters.
• Assume P(t) and g(t) are continuous on , and let
(t) be a fundamental matrix for the homogeneous system.
• Recall that the columns of are linearly independent
solutions of x' = P(t)x, and hence (t) is invertible on the
interval , and also .
• Next, recall that the solution of the homogeneous system
can be expressed as x = (t)c.
• Analogous to Section 3.7, assume the particular solution of
the nonhomogeneous system has the form x = (t)u(t),
where u(t) is a vector to be found.
a < t < b
Y
YY
a < t < b Y '(t) = P(t)Y(t)
Y
Y
Variation of Parameters: Solution
• We assume a particular solution of the form x = (t)u(t).
• Substituting this into x' = P(t)x + g(t), we obtain
'(t)u(t) + (t)u'(t) = P(t) (t)u(t) + g(t)
• Since '(t) = P(t) (t), the above equation simplifies to
u'(t) = –1(t)g(t)
• Thus
where the vector c is an arbitrary constant of integration.
• The general solution to x' = P(t)x + g(t) is therefore
cΨu dttgtt )()()( 1
arbitrary,,)()()()( 1
1
1
tdssstt
t
tgΨΨcΨx
Y
Y Y Y
Y Y
Y
Variation of Parameters: Initial Value
Problem
• For an initial value problem
x' = P(t)x + g(t), x(t0) = x(0),
the general solution to x' = P(t)x + g(t) is
• Alternatively, recall that the fundamental matrix (t)
satisfies (t0) = I, and hence the general solution is
• In practice, it may be easier to row reduce matrices and
solve necessary equations than to compute –1(t) and
substitute into equations. See next example.
t
tdsssttt
0
)()()()()( 1)0(
0
1gΨΨxΨΨx
t
tdssstt
0
)()()()( 1)0(gΨΦxΦx
FF
Y
Example 3: Variation of Parameters (1 of 3)
• Consider again the nonhomogeneous system x' = Ax + g:
• We have previously found general solution to homogeneous
case, with corresponding fundamental matrix:
• Using variation of parameters method, our solution is given
by x = (t)u(t), where u(t) satisfies (t)u'(t) = g(t), or
¢x =-2 1
1 -2
æ
èç
ö
ø÷x+
2e-t
3t
æ
èç
ö
ø÷ =
-2 1
1 -2
æ
èç
ö
ø÷x+
2
0
æ
èç
ö
ø÷e-t +
0
3
æ
èç
ö
ø÷t
tt
tt
ee
eet
3
3
)(Ψ
t
e
u
u
ee
ee t
tt
tt
3
2
2
1
3
3
Y Y
Example 3: Solving for u(t) (2 of 3)
• Solving (t)u'(t) = g(t) by row reduction,
• It follows that
2/31
2/3
2/3110
2/301
2/30
2/30
2/30
2
3220
2
3
2
2
32
1
32
33
3
3
3
t
tt
t
tt
tt
tt
tt
ttt
tt
ttt
tt
ttt
teu
teeu
te
tee
tee
tee
tee
eee
tee
eee
tee
eee
u(t) =u1
u2
æ
èç
ö
ø÷ =
e2t / 2 - te3t / 2 + e3t / 6 + c1
t + 3tet / 2 - 3et / 2 + c2
æ
è
çç
ö
ø
÷÷
Y
Example 3: Solving for x(t) (3 of 3)
• Now x(t) = (t)u(t), and hence we multiply
to obtain, after collecting terms and simplifying,
• Note that this is the same solution as in Example 1.
2
1
332
3
3
2/32/3
6/2/2/
cetet
cetee
ee
eett
ttt
tt
tt
x
x = c1
1
-1
æ
èç
ö
ø÷e-3t + c2
1
1
æ
èç
ö
ø÷e- t +
1
1
æ
èç
ö
ø÷te-t +
1
2
1
-1
æ
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ö
ø÷e-t +
1
2
æ
èç
ö
ø÷t -
1
3
4
5
æ
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ö
ø÷
Y
Laplace Transforms
• The Laplace transform can be used to solve systems of
equations. Here, the transform of a vector is the vector of
component transforms, denoted by X(s):
• Then by extending Theorem 6.2.1, we obtain
)()(
)(
)(
)()(
2
1
2
1s
txL
txL
tx
txLtL Xx
)0()()( xXx sstL
Example 4: Laplace Transform (1 of 5)
• Consider again the nonhomogeneous system x' = Ax + g:
• Taking the Laplace transform of each term, we obtain
where G(s) is the transform of g(t), and is given by
t
e t
3
2
21
12xx
G(s) =
2
s +1
3
s2
æ
è
çççç
ö
ø
÷÷÷÷
)()()0()( ssss GAXxX
Example 4: Transfer Matrix (2 of 5)
• Our transformed equation is
• If we take x(0) = 0, then the above equation becomes
or
• Solving for X(s), we obtain
• The matrix (sI – A)–1 is called the transfer matrix.
)()()0()( ssss GAXxX
)()()( ssss GAXX
)()(1
sss GAIX
)()( sss GXAI
Example 4: Finding Transfer Matrix (3 of 5)
• Then
• Solving for (sI – A)–1, we obtain
21
12
21
12
s
ss AIA
21
12
)3)(1(
11
s
s
sss AI
Example 4: Transfer Matrix (4 of 5)
• Next, X(s) = (sI – A)–1G(s), and hence
or
23
)1(2
21
12
)3)(1(
1)(
s
s
s
s
sssX
)3(1
)2(3
)3(1
2
)3(1
3
)3(1
22
)(
22
22
sss
s
ss
sssss
s
sX
Example 4: Transfer Matrix (5 of 5)
• Thus
• To solve for x(t) = L-1{X(s)}, use partial fraction expansions
of both components of X(s), and then Table 6.2.1 to obtain:
• Since we assumed x(0) = 0, this solution differs slightly
from the previous particular solutions.
)3(1
)2(3
)3(1
2
)3(1
3
)3(1
22
)(
22
22
sss
s
ss
sssss
s
sX
x = -2
3
1
-1
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ö
ø÷e-3t +
2
1
æ
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ö
ø÷e-t +
1
1
æ
èç
ö
ø÷te-t +
1
2
æ
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ö
ø÷t -
1
3
4
5
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Summary (1 of 2)
• The method of undetermined coefficients requires no
integration but is limited in scope and may involve several
sets of algebraic equations.
• Diagonalization requires finding inverse of transformation
matrix and solving uncoupled first order linear equations.
When coefficient matrix is Hermitian, the inverse of
transformation matrix can be found without calculation,
which is very helpful for large systems.
• The Laplace transform method involves matrix inversion,
matrix multiplication, and inverse transforms. This method
is particularly useful for problems with discontinuous or
impulsive forcing functions.
Summary (2 of 2)
• Variation of parameters is the most general method, but it
involves solving linear algebraic equations with variable
coefficients, integration, and matrix multiplication, and
hence may be the most computationally complicated
method.
• For many small systems with constant coefficients, all of
these methods work well, and there may be little reason to
select one over another.
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