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MA1506

Mathematics II

Chapter 7Systems of First Order ODEs

1Chew T S MA1506-14 Chapter 7

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7.1 Solving Linear System of ODEs

a,b,c,d constantsHow to solve

i.e,

We shall look at an old problem,

which is related to our new problem,to get ideas to solve our new problem


dxax by

dt = + dy cx dy

dt = +

x a b xd y c d ydt

=

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An old problemConsider ( )

2 ( )dx t

x t dt =

Zero function ( ) 0 x t is a solutionHowever we are interested in nonzero solutions

We know that2

0( ) t x t x e=

is the general solution , where 0 x is any constantIf we want nonzero solutions, then we assumethat 0 x is nonzero constant

7.1 Solving System of ODEs


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So solutions of the old problem are of theform 0( )

t x t x e =

From this old problem, we may guess thatsolutions of

are of the form 0( ) t x t x e = 0( )

t y t y e =

which can be written as 00

( )( )

t x x t e y y t

=

Solutions of systems of ODE



x a b xd y c d ydt

=

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Now we shall find 0 0, , x y

How to find? First substitute

0

0

( )

( )t x x t e

y y t

=

into the given ODE,

Note that



0

0

( )

( )t x x t d e

y y t dt

=

x a b xd y c d ydt

=

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0 0

0 0

t t x xe B e y y

=

wherea b

Bc d

=

Let 000

xu

y

=

We get

0 0u Bu =number 2x2 matrix

We can not apply cancellation rule to the above equality



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0 0u Bu =Rewrite

as 2 0 0 I u Bu = where 21 0

0 1 I =



Hence we have 2 0( ) 0 B I u =

We want nonzero solns of the given ODE,so we want0u to be nonzero vector

2det( ) 0 B I =

from Chapter 5, Section 5.5.

Therefore

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Before we look at

2det( ) 0 B I =

2 0( ) 0 B I u =

We shall look at the following equality carefully

This equality can be written as

0 0 Bu u =eigenvector eigenvalue

So we are looking for eigenvaluesand eigenvectors of B



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2det( ) deta b

B I c d

= 0=

Hence ( )( ) 0a d bc =

Thus 21 ( ) 4( )2

a d a d ad bc = + +

[ ] [ ] [ ]21 ( ) 4(det )2

Tr B Tr B B =

Recall a b Bc d

=

2 ( ) ( ) 0a d ad bc + + =

Eigenvalue



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There are three casesCase 1 : Two distinct real roots (eigenvalues)

i.e.,

Suppose two distinct real eigenvalues are 1 2,

Then find the corresponding eigenvectors 1 2,u uby

1 2 1( ) 0 B I u = 2 2 2( ) 0 B I u =

1

1t u e 2

2t u e and

We have two (linearly independent) solutions

[ ]( ) [ ]2 4Tr B Det B>



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The general solution is

1 2

1 1 2 2t t c u e c u e +

where 1 2,c c are any real numbers

..(C1)



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Case 2 : Two complex roots (eigenvalues)

is given by

i.e., [ ]( ) [ ]2 4Tr B Det B

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Hence for Case 1:

Zero equilibrium soln is stable iff

and det( ) 0 B

Note: In fact det( ) 0 B >

since we assume that inverse of B exits, so

det( ) 0 B

( ) 0Tr B

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Case 2: Two complex eigenvalues

The general soln is

[ ][ ]

1

2

cos sin

sin cos

t

t

c e u t v t

c e u t v t

+ +

where1

( )2 Tr B =(See Case 2 in 7.1)

7.2 Stability of equilibrium


2( ) 4detTrB B

i.e.,0 ( ) 0Tr B

7.2 Stability of equilibrium


2( ) 4detTrB B

det B

TrBFor case1 ( ) 0Tr B 0 : Nodal source

Both < 0 : Nodal sink

Two Real Eigenvalues

Next we shall use ( ),det( )Tr B Bto classify the above three cases,which is easier

p


Classification of zero solution

7.3 Phase plane

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Eigenvalues are given by

Real roots

Nodal Source Both > 0

Nodal Sink Both < 0

Saddle Opp Signs

Recall: we do not discuss zero eigenvector

p


[ ] [ ] [ ]21 ( ) 4(det )2

Tr B Tr B B =

2( ) 4detTrB B>

0TrB >0TrB

det 0 B >

det 0 B

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7.5. WarfareSuppose G(0)=15000 is fixed

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G

M

M will win

G will win All will be killed


23

M G=

15000

9999

1500010001

pp ( )Then min M(0) that M willwin is 10001

7.5. Warfare

Suppose M(0)=10000 is fixed

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G

M

M will win

G will win All will be killed

Suppose M(0)=10000 is fixedThen min G(0) that G will winis 15001

2

3 M G=

1500110000

14999

10000

EndChapter 7

chewma1506-14 ch7

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