Transcript Math 260
Ch 7.5: Homogeneous Linear Systems with
Constant Coefficients
We consider here a homogeneous system of n first order linear
equations with constant, real coefficients:
x1 a11 x1 a12 x2 a1n xn
x2 a21 x1 a22 x2 a2 n xn
xn an1 x1 an 2 x2 ann xn
This system can be written as x' = Ax, where
x1 (t )
a11
x2 (t )
a21
x(t )
, A
x (t )
a
m
n1
a12
a22
an 2
a1n
a2 n
ann
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
rξert Aξert
rξ Aξ
A rIξ 0
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.
Example 1: Direction Field
(1 of 9)
Consider the homogeneous equation x' = Ax below.
1 1
x
x
4 1
A direction field for this system is given below.
Substituting x = ert in for x, and rewriting system as
(A-rI) = 0, we obtain
1 1 0
1 r
4 1 r 1 0
Example 1: Eigenvalues (2 of 9)
Our solution has the form x = ert, where r and are found
by solving
1 1 0
1 r
4 1 r 1 0
Recalling that this is an eigenvalue problem, we determine r
by solving det(A-rI) = 0:
1 r
1
(1 r ) 2 4 r 2 2r 3 (r 3)(r 1)
4 1 r
Thus r1 = 3 and r2 = -1.
Example 1: First Eigenvector (3 of 9)
Eigenvector for r1 = 3: Solve
1 1 0
1 3
A rI ξ 0
4 1 3 2 0
1 1 0
2
4 2 2 0
by row reducing the augmented matrix:
1 0 1 1/ 2
2
2
4 2 0 4
1 / 2 2
1 / 2
(1)
c
, c
ξ
1
2
1 1 / 2 2
0 1 1/ 2 0
1
0 2
0 0
0 0
1
(1)
arbitrary choose ξ
2
0
0
Example 1: Second Eigenvector (4 of 9)
Eigenvector for r2 = -1: Solve
1 1 0
1 1
A rI ξ 0
4 1 1 2 0
2 1 1 0
4 2 2 0
by row reducing the augmented matrix:
1 1 / 2 2 0
2 1 0 1 1/ 2 0 1 1/ 2 0
1
0 2 0
2 0 0
0 0
4 2 0 4
1 / 2 2
1/ 2
1
( 2)
( 2)
c
, c arbitrary choose ξ
ξ
2 1
2
Example 1: General Solution (5 of 9)
The corresponding solutions x = ert of x' = Ax are
1 3t ( 2)
1 t
x (t ) e , x (t ) e
2
2
(1)
The Wronskian of these two solutions is
W x (1) , x ( 2)
e 3t
(t ) 3t
2e
e t
2t
4
e
0
t
2e
Thus x(1) and x(2) are fundamental solutions, and the general
solution of x' = Ax is
x(t ) c1x (1) (t ) c2 x ( 2 ) (t )
1 3 t
1 t
c1 e c2 e
2
2
Example 2:
(1 of 9)
Consider the homogeneous equation x' = Ax below.
3
2
x
x
2 2
Substituting x = ert in for x, and rewriting system as
(A-rI) = 0, we obtain
3 r
2
2 1 0
2 r 1 0
Example 2: Eigenvalues (2 of 9)
Our solution has the form x = ert, where r and are found
by solving
3 r
2
2 1 0
2 r 1 0
Recalling that this is an eigenvalue problem, we determine r
by solving det(A-rI) = 0:
3 r
2
2
2r
(3 r )(2 r ) 2 r 2 5r 4 (r 1)(r 4)
Thus r1 = -1 and r2 = -4.
Example 2: First Eigenvector (3 of 9)
Eigenvector for r1 = -1: Solve
3 1
2 1 0
A rI ξ 0
2
2
1
2 0
2
2
2 1 0
1 2 0
by row reducing the augmented matrix:
2
2
ξ (1)
2 0 1 2 / 2 0 1 2 / 2 0
1 0 2
1 0 0
0 0
2 / 2 2
1
(1)
choose ξ
2
2
Example 2: Second Eigenvector (4 of 9)
Eigenvector for r2 = -4: Solve
3 4
2 1 0
A rI ξ 0
2 2 4 2 0
1
2
by row reducing the augmented matrix:
1
2
2 0 1
2 0 0
choose ξ
( 2)
2
1
2 2
2 0
ξ ( 2)
0 0
2
2 1 0
2 2 0
Example 2: General Solution (5 of 9)
The corresponding solutions x = ert of x' = Ax are
2 4t
1 t ( 2)
e
x (t ) e , x (t )
1
2
(1)
The Wronskian of these two solutions is
W x (1) , x ( 2 ) (t )
e t
2e 4t
2e t
e 4t
3e 5t 0
Thus x(1) and x(2) are fundamental solutions, and the general
solution of x' = Ax is
x(t ) c1x (1) (t ) c2 x ( 2) (t )
2 4t
1 t
e
c1 e c2
1
2
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
x(1) (t ) ξ (1)er1t ,, x( n) (t ) ξ ( n)ernt
Using Wronskian, it can be shown that these solutions are
linearly independent, and hence form a fundamental set of
solutions. Thus general solution is
x c1ξ (1)er1t cnξ ( n)ernt
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
x(1) (t ) ξ (1)er1t ,, x( n) (t ) ξ ( n)ernt
and form a fundamental set of solutions.
Example 3: Hermitian Matrix
(1 of 3)
Consider the homogeneous equation x' = Ax below.
0 1 1
x 1 0 1 x
1 1 0
The eigenvalues were found previously in Ch 7.3, and were:
r1 = 2, r2 = -1 and r3 = -1.
Corresponding eigenvectors:
ξ (1)
1
1
0
( 2 ) ( 3)
1, ξ 0 , ξ 1
1
1
1
Example 3: General Solution (2 of 3)
The fundamental solutions are
x (1)
1
1
0
2 t ( 2 ) t ( 3) t
1 e , x 0 e , x 1 e
1
1
1
with general solution
1
1
0
2t
t
t
x c1 1e c2 0 e c3 1e
1
1
1