Matrix Methods for Solving Systems of 1st Order Linear ...

Matrix Methods for Solving Systems of 1st Order Linear Differential Equations

The Main Idea:

Given

a system of 1st order

linear differential equations

dx dt

= Ax

with initial conditions

x(0) , we use

eigenvalue-eigenvector analysis to find an appropriate basis B = {v1,, vn} for Rn and a change of basis

matrix

S

=

v1

v

n

such

that in

coordinates

relative

to

this

basis

(u

= S-1x )

the

system is

in

a

standard

form with a known solution. Specifically, we find a standard matri= x B [A= ]B S-1AS , transform the system

into

du dt

= Bu , solve it as

u(t) = [etB ]u(0)

where

[etB ]

is the evolution matrix for B, then transform back to the

original coordinates to get x(t) = [etA ]x(0) where [etA ] = S[etB ]S-1 is the evolution matrix for B. That is

x= (t) [e= tA ] S[etB ]S-1x(0) . This is actually easier to do than it is to explain, so here are a few illustrative

examples:

The diagonalizable case

Problem:

Solve

the

system

d=x dt d=y

5x

-

6

y

with initial

3x - 4y

dt

conditions= x(0) 3= , y(0) 1.

Solution: In matrix form, we have

dx dt

= Ax

where

A

=

5 3

-6 -4

and

x(0)

=

3 1

.

We

start

by

finding

the

eigenvalues of the matrix: I - A =--35

6 +

4

,

and

the

characteristic polynomial is pA () = 2 - - 2 = ( - 2)( +1) . This gives the eigenvalues 1 = 2 and 2 = -1

.

The

first

of these

gives

the

eigenvector

v1

=

2 1

,

and

the second

gives

the eigenvector

v2

=

1 1

.

So we

have

AAvv21

= =

1v1 2v2

.

The

change

of

basis

matrix

is

S

=

2 1

1 1

and with the new basis (of eigenvectors)

B = {v1, v2} we have = [A]B S= -1AS 01= 02 02= -01 D , a diagonal matrix. [There is no need to carry

out the multiplication of the matrices if B = {v1, v2} is known to be is a basis of eigenvectors. It will always yield a diagonal matrix with the eigenvalues on the diagonal.]

The

evolution

matrix

for

this

diagonal

matrix

is

[etD ]

=

e2t

0

0 e-t

,

and

the

solution

of

the

system

is:

= x(t)

[e= tA ]x(0)

S[etD ]= S-1x(0)

2 1 e2t 1 1 0

0 1 e-t -1

= -21 13

2e2t e2t

e-t 2 e-t -1

=

4e2t 2e2t

- -

e-t e-t

=

2e2t

2 1

-

e-t

1 1

=

2e2t v1

-

e-t v2

1

Revised May 9, 2017

The complex eigenvalue case

Let A be a matrix with a complex conjugate pair of eigenvalues = a + ib and = a - ib . We can proceed just as in the case of real eigenvalues and find a complex vector w such that (I - A)w =0 . The components of such

a vector w will have complex numbers for its components. If we decompose w into its real and imaginary vector components as w= u + iv (where u and v and real vectors), we can calculate that:

(1)

Aw =Au + iAv =w =(a + ib)(u + iv) =(au - bv) + i(bu + av)

If we define the vector w^= u - iv and use the easy-to-prove fact that for a matrix A with all real entries we'll have Aw = Aw^ = w = w^ , we see that w^= u - iv will also be an eigenvector with eigenvalue , and:

(2)

Au - iAv = (au - bv) - i(bu + av)

The true value of this excursion into the world of complex numbers and complex vectors is seen when we add and subtract equation (1) and (2). We get:

2= Au 2(au - bv) 2= iAv 2i(bu + av)

After cancellation of the factors of 2 and 2i in the respective equations and rearranging, we get:

A=v av + bu Au =-bv + au

Note that we are now back in the "real world": all vectors and scalars in the above equations are real. If we use

the two vectors B = {v,u} as basis vectors associated with the two complex conjugate eigenvalues, we see that

in coordinates associated with this basis (and change of basis matrix S = [v u] ) we'll have the matrix [A]B of

the form:

a

-b

[A] B

= S-1AS = ba -ab = a2 + b2 b aa

2 2

+ b2 + b2

a

a2 a2

+ b2 + b2

= csoins -csoisn

= R

where R is the rotation matrix corresponding to the angle= arg() .

Next, we want to find the evolution matrix for this (real) normal form.

In

fact,

[etB ]

=

eat

cos bt sin bt

- sin bt cos bt

,

a

time-varying

rotation

matrix

with

exponential

scaling.

This

yields

a

trajectory that spirals out in the case where Re()= a > 0 (look to the original vector field to see whether it's

clockwise or counterclockwise), or a trajectory that spirals inward toward 0 in the case where Re()= a < 0 .

To derive this expression for [etB ] , make another coordinate change with complex eigenvectors starting

with

B

=

a b

-b a

.

B

will

have

the

same

eigenvalues

of

A,

namely

=

a + ib and =

a - ib , and

I - B =--ba

b -

a

.

Using

the

eigenvalue

=

a + ib ,

we

seek

a

complex

eigenvector

such

that

-ib= b ibb b

= -ibb ++ bibbb

0 0

.

This

implies

that

=

i ,

so

one

such

complex

eigenvector

is

w

=

i 1

.

The

2

Revised May 9, 2017

eigenvalue =

a - ib

will

then

give

eigenvector

w^

=

-i 1

.

Using

the

(complex)

change

of

basis

matrix

P

=

i 1

-i 1

,

we

have

that

P-1BP=

D=

a + ib 0

a

0 - ib

.

Just

as

in

the

case

of

real

eigenvalues,

it

follows

that:

= [etB ]

P= [etD ]P-1

1 i 2i 1

-i e(a+ib)t

1

0

e(

0

a -ib ) t

= -11 ii

eibt +e-ibt

eat

eibt

2

-e-ibt 2i

-eiebi= tb+t2-2ee-i -ibitbt

eat

cos bt sin bt

- sin bt cos bt

.

Note that the exponential factor eat will grow if a= Re() > 0 and decay if a= Re() < 0 . Further note that

the

matrix

cos bt sin bt

- sin bt cos bt

is

a

time-varying

rotation

matrix

with

rotational

frequency

b.

The

product

of

the

exponential factor and the time-varying rotation matrix means that the trajectories associated with the evolution

matrix [etB ] will be either outward or inward spirals depending upon whether a > 0 or a < 0 . In the case where

a = 0 we would get closed (periodic) trajectories ? circles, in fact, for this standard case.

These calculations enable us to write down a closed form expression for the solution of this linear system,

namely x(t) = [etA ]x(0) whe= re [etA ]

S= [etB ]S-1

eat

S

cos sin

bt bt

- sin bt cos bt

S-1

.

However,

the

more

important

result

is the ability to qualitatively describe the trajectories for this system by knowing only the real part of the

eigenvalues of the matrix A and the direction of the corresponding vector field (clockwise vs.

counterclockwise).

Problem:

Solve

the

system

d=x dt d=y

2x

-

5

y

with initial conditions

2x - 4y

dt

= x(0) 0= , y(0) 1.

Solution:

In

matrix

form,

we

have

dx dt

=

Ax

where

A

=

2 2

-5 -4

and

x(0)

=

0 1

.

We

again

start

by

finding

the

eigenvalues

of

the

matrix: I - A =--22

5 +

4

,

and

the

characteristic

polynomial

is pA () = 2 + 2 + 2 = ( +1)2 +1. This gives the complex

eigenvalue pair = -1+ i and = -1- i . We seek a complex eigenvector for the first of these:

-3 + i -2

5 3 + i

=

0 0

gives

the

(redundant)

equations

(-3 + i)

+ 5

= 0

and

-2

+ (3 + i)

=0 .

The

first

of these can be written as 5= (3 - i) , and an easy solution to this is where = 5, = 3 - i . (We could also

have used the second equation ? which is a scalar multiple of the first. The eigenvector might then have been different, but ultimately we'll get the same result.) This gives the complex eigenvector

w

= 3 5-

i

= 53 + i

0 -1

= u + iv

.

We

have

shown

that

with

the

specially

chosen

basis

B

= {v,u} ,

the

new

system will have standard matrix = [A]B S= -1AS ba= -ab B where a is the real part of the complex

3

Revised May 9, 2017

eigenvalue

and

b

is

its

imaginary

part.

We

also

showed

that

[etB ]

=

eat

cos bt sin bt

- sin bt cos bt

.

In

this

example,

a = -1 and b == 1, S

[= v u]

0 -1

5 3

,

S-1

=

1 5

3 1

-5 0

,

B

=

-1 1

-1 -1

,

and

[etB ]

=

e-t

cos t sin t

- sin t cos t

.

The

solution to the system is therefore x= (t) [e= tA ] S[etB ]S-1x= (0)

e-t 0 5 -1

5 cos t 3 sin t

- sin t 3 cos t 1

-5 0 0 1

e= 5-t - cos5tsi+n3tsin t sin5t +co3sctos t -05

e-t

-5sin t cos t - 3sin

t

.

That

is= , yx((tt)) =

-5e-t sin t e-t (cos t -

3

sin

t

)

.

Repeated eigenvalue case [with geometric multiplicity (GM) less than the algebraic multiplicity (AM)]:

Problem:

Solve

the

system

dx

dt dy

= =-4x

y

+ 4y

with

initial

dt

conditions= x(0) 3= , y(0) 2 .

Solution: In matrix form, we have

dx dt

=

Ax

where

A

=

0 -4

1 4

and

x(0)

=

3 2

.

We

again

start

by

finding

the

eigenvalues

of

the

matrix: I - A =4

-1 - 4

,

and

the

characteristic

polynomial

is

pA () = 2 - 4 + 4 = ( - 2)2 . This gives the repeated eigenvalue =2 with (algebraic) multiplicity 2. We

seek

eigenvectors:

2 4

-1 -2

=

0 0

gives

the

(redundant)

equations

2 - =0

and

4 - 2 =0 .

Therefore

=

2 , so we can choose

v 1

=

1 2

or any scalar multiple of this as an eigenvector, but we are unable to find

a second linearly independent eigenvector. (We say that the geometric multiplicity of the =2 eigenvalue

is 1.)

The standard procedure in this case is to seek a generalized eigenvector for this repeated eigenvalue, i.e. a

vector v2 such that (I - A)v2 is not zero, but rather a multiple of the eigenvector v1 . Specifically, we seek a

vector such that Av2= v1 + v2 . This translates into seeking v2 such that (I - A)v2 =-v1 . That is,

2 4

-1 -2

=

-1 -2

.

This

gives

redundant

equations

the

first

of

which

is

2 -

=

- 1

or

=

2 +1. If we

(arbitrarily)

choose

=0 ,

then

=1,

so

v2

=

0 1

.

The

fact

that

AAvv=21 =

2v1 v1 +

2

v

2

tells us

that

with the

change

of

basis

matrix

S

=

1 2

0 1

,

we

will

have

= [A]B

S= -1AS

= 02 12

B.

The

standard

form

in

this

repeated

eigenvalue

case

is

a

matrix

of

the

form

B

=

0

1

.

(There

are

analogous

forms in cases larger than 2? 2 matrices.) Note that we can write B =0

1

=I

+P

where

P

=

0 0

1 0

.

4

Revised May 9, 2017

There is a simple relationship between the solutions of the systems

dx dt

= Bx

and

du dt

= Pu , namely

x(t) = etu(t) . This is easily seen by differentiation:

dx dt

=

d dt

[et

u(t

)]

=

et

du dt

+ etu

=

etPu + etu =

et (Pu + Iu) =

et (I + P)u =

(I + P)etu =

Bx

together with the fact that

x(0) = u(0) . Furthermore, solving

du dt

= Pu

is simple. If

u

=

u1 u2

,

then

with

the

matrix

P

=

0 0

1 0

we

have

uu21

(t (t

) )

= =

u2 0

.

The

second equation

gives

that

u2 (t=)

c=2

u2 (0) , a constant. The

first equation is then u1(t) = u2 (0) , so u1= (t) u2 (0) t + c1 . At t = 0 this gives u1(0) = c1 , so

u1(t) = u1(0) + u2 (0) t . Together this gives:

= u(t)

= uu12((tt))

u1

(0)

+

uu22= ((00))t

1 0

t 1

uu= 12((00))

1 0

1t= u(0)

etP u(0)

T= herefore x(t)

e= t 10 1t x(0)

et

0

tet et

x(0)

,

so

etB

=

et

0

tet

et

for

B

=

0

1

.

If

we

apply

this

to

the

problem

at

hand,

we

get

[etB ]

=

e2t

0

te2t e2t

.

The

solution

to

the

system

is

therefore

x= (t) [e= tA ]

S[etB ]S-1x= (0)

1 2

0 e2t

1

0

te2t 1

e2t

-2

0 1

= 32

e2t

2e2t

te2t 3

2te2t

+

e2t

-4

6e23t e-2= 8t t-e24tte-24t e2t

= 32ee22tt -- 48ttee22tt

e2t

3 - 4t 2 - 8t

.

That

is,

= x(t) = y(t)

e2t e2t

(3 (2

- -

4t 8t

) )

.

It's worth noting that this can also be expressed as= x(t)

e2t

3 2

-

4te2t

1 2

.

The phase portrait in this case has just one invariant (eigenvector) direction. It gives an unstable node which can be viewed as a degenerate case of a (clockwise) outward spiral that cannot get past the eigenvector direction.

Moral of the Story: It's always possible to find a special basis relative to which a given linear system is in its simplest possible form. The new basis provides a way to decompose the given problem into several simple, standard problems which can be easily solved. Any complication in the algebraic expressions for the solution is the result of changing back to the original coordinates.

The standard 2? 2 cases are:

Diagonalizable with eigenvalues 1, 2 :

B= D=

1 0

0 2

Complex pair of eigenvalues = a ? ib :

B

=

a b

-b a

Repeated eigenvalue with GM < AM :

B

=

0

1

[e= tB ]

[e= tD ]

e1t

0

0

e

2t

[etB

]

=

eat

cos bt sin bt

- sin bt cos bt

[etB

]

=

et

0

tet

et

5

Revised May 9, 2017

In general, you should expect to encounter systems more complicated than these 2? 2 examples. To illustrate the line of reasoning in a significantly more complicated case, here is a Big Problem.

Big Problem: a) Find the general solution for the following system of differential equations:

dx1

dt dx2

dt

=

2 x1

-

4 x4

+

3x5

=

2 x2

-

2 x3

+

2 x4

dx=3

dt

x2 - x4

dx4 dt

=

- x4

dx5

dt

= -3x4

+ 2x5

5 4 b) Find the solution in the case where x(0) = 3 . 2 1

2 0 0 -4 3

Solution: This is

a continuous dynamical

system of the form

dx

= Ax

where

0 A = 0

2 1

-2 0

2 -1

0 0 .

dt

0 0 0 -1 0

0 0 0 -3 2

- 2 0 0 4 -3 0 - 2 2 -2 0

We start by seeking the eigenvalues. We have I - A = 0 -1 1 0 .

0 0 0 +1 0 0 0 0 3 - 2

The characteristic polynomial is pA () = ( - 2)2 ( +1)(2 - 2 + 2) which yields the repeated eigenvalue 1 =2 =2 (with algebraic multiplicity 2), the distinct eigenvalue 3 =-1 , and the complex pair 4 = 1+ i and 5 =4 =1- i .

1 0 The repeated eigenvalue 1 =2 =2 yields just one eigenvector v1 = 0 , so its geometric multiplicity if just 1. 0 0

We then seek a "generalized eigenvector" v2 such that Av2= v1 + v2 where =2 . That is, we seek a vector

v2 such that v2 - Av2 =(I - A)v2 =-v1 . This is just an inhomogeneous system which yields solutions of the

t

0

0

0

form v2 = 0 . For simplicity, take the solution with t = 0 , i.e. v2 = 0 .

0

0

1 3

1 3

6

Revised May 9, 2017

1 0 The eigenvalue 3 =-1 yields the eigenvector v3 = 3 . A straightforward calculation with the complex 3 3

0 0 0 1+ i 1 1 eigenvalue 4 = 1+ i yields the complex eigenvector v = 1 =1 + i 0 =v5 + i v4 in accordance with the 0 0 0 0 0 0

method previously derived.

1 0 1 0 0

Using the ba= sis B= v1

0 = 0 , v2

0 = 0 , v3

0 = 3 , v4

1 = 0 , v5

1 1

and

change

of

basis

matrix

0 0 3 0 0

0

13

3

0

0

1 0 1 0 0 0 0 0 1 1

1 0

0

-

1 3

0

0 0 0 -3 3

S = 0 0 3 0 1 , we compute the inverse matrix S-1 = 0 0 0

1 3

0 .

0 0 3 0 0

0 1 -1 1 0

0

1 3

3

0

0

0 0 1 -1 0

Av1 = 2v1

We know tha= t AA= vv23

v1 + 2v2 - v3

,

so

the

matrix

of

A

relative

to

the

basis

B

is

= Av4

v4 + v5

= Av 5

- v4 + v5

2 1 0 0 0 0 2 0 0 0 =B S= -1AS 0 0 -1 0 0 .

0 0 0 1 -1 0 0 0 1 1

Since A = SBS-1 , it will be the case that the evolution matrices are related via etA = S etB S-1 where

e2t te2t 0 0

0

0 e2t 0 0

0

etB = 00

0 0

0 0

e-t 0 0

0 et cos t et sin t

0 -et sin et cos t

t

.

The solution is then

= x(t)

= etA x(0)

S etB= S-1x(0)

1 0

0 0 0

0 0 0 0

1 3

1 0 3 3 3

0 1 0 0 0

0 e2t

1 1 0 0

0 00 0

te2t e2t 0 0 0

0 0 e-t 0 0

0

0

0 et cos t et sin t

0

0

0 -et sin et cos t

t

1 0 0 0 0

0 0 0 1 0

0

-

1 3

0 -3

0

1 3

-1 1

1 -1

0 3 0 x(0) . 0

0

7

Revised May 9, 2017

c1

If

we

multiply

the

leftmost

matrices

and

write

S-1x(0)

=

c2

ccc534

,

this

yields

the

general

solution:

= x(t)

= etA x(0)

S etB= S-1x(0)

e2t

0

0 0

0

te2t 0 0 0 e1 2t

3

e-t

0 3e-t 3e-t 3e-t

0 et (cos t + sin t)

et sin t

0

0

et

0 (cos t - sin

et cos t

0

0

t

)

c1 c2 ccc534

or

x1 x2 x3

(t) =c1e2t + c2te2t + c3e-t (=t) c4et (cos t + sin t) + c5et (t) =3c3e-t + c4et sin t + c5et

(cos t cos t

-

sin

t)

.

= xx45 ((tt)) =

3c3e-t

1 3

c2

e2t

+ 3c3e-t

5 4 If, on the other hand, we use the initial condition x(0) = 3 , we get the specific solution:

2 1

e2t

0

x(t)

=

0 0

0

te2t 0 0 0 e1 2t

3

e-t

0 3e-t 3e-t 3e-t

0 et (cos t + sin t)

et sin t

0

0

et

0 (cos t - sin

et cos t

0

0

t

)

133 -3 23 3 1

or= xxx231(((ttt))) == e2133te(e-42t tc+-oes3ttt(e3+2st2i+nsit23n+et-)ct os

t

)

.

x4 x5

(t (t

) )

= 2e-t =-e2t

+

2e-t

8

Revised May 9, 2017

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