Linear Systems of ODE: Solving the IVP - Clemson...

Calculus For Biologists

Linear Systems of ODE: Solving the IVP

James K. Peterson

Department of Biological Sciences and Department of Mathematical SciencesClemson University

October 6, 2013


Outline

1 Linear Systems of ODE

2 The Characteristic EquationExamples

3 Finding The General SolutionExamples


Abstract

This lecture is about linear systems of ODE and how to solve theresulting IVP.


Linear Systems of ODE

We are now ready to solve what are called Linear Systems of firstorder ODE. These have the form

x ′(t) = a x(t) + b y(t)

y ′(t) = c x(t) + d y(t)

x(0) = x0

y(0) = y0

for any numbers a, b, c and d and initial conditions x0 and y0.The full problem is called, as usual, an Initial Value Problem orIVP for short. The two initial conditions are just called the IC’s forthe problem to save writing.


Linear Systems of ODE

For example, we might be interested in the system

x ′(t) = −2 x(t) + 3 y(t)

y ′(t) = 4 x(t) + 5 y(t)

x(0) = 5

y(0) = −3

Here the IC’s are x(0) = 5 and y(0) = −3.

Another sample problem might be the one below.

x ′(t) = 14 x(t) + 5 y(t)

y ′(t) = −4 x(t) + 8 y(t)

x(0) = 2

y(0) = 7


The Characteristic Equation

For linear first order problems like u′ = 3u and so forth, wefind the solution has the form u(t) = A e3t for some numberA. We then determine the value of A to use by looking at theinitial condition.

To find the solutions here, we begin by rewriting the model inmatrix - vector notation.[

x ′(t)y ′(t)

]=

[a bc d

] [x(t)y(t)

].

The 2× 2 matrix is called the coefficient matrix of thismodel.

The initial conditions can then be redone in vector form as[x(0)y(0)

]=

[x0y0

].



Now it seems reasonable to believe that if a constant times ert

solves a first order linear problem like u′ = ru, perhaps a vectortimes ert will work here. Let’s make this formal. So let’s look atthe problem below

x ′(t) = 3 x(t) + 2 y(t)

y ′(t) = −4 x(t) + 5 y(t)

x(0) = 2

y(0) = −3

Assume the solution has the form V ert

Let’s denote the components of V as follows:

V =

[V1

V2

].



We assume the solution is[x(t)y(t)

]= V ert .

Then the derivative of V ert is(V ert

)′=

([V1

V2

]ert)′

=

([V1 ert

V2 ert

] )′=

[V1 (ert)′

V2 (ert)′

]=

[V1 r ert

V2 r ert

]=

[V1

V2

]r ert = r V ert

Hence,

[x ′(t)y ′(t)

]= r V ert .



When we plug these terms into the matrix - vector form of theproblem, we find

r V ert =

[3 2−4 5

]V ert

Rewrite as

r V ert −[

3 2−4 5

]V ert =

[00

].

Recall that the 2× 2 identity matrix I has the form

I =

[1 00 1

]⇒ I V = V.



So

r V ert −[

3 2−4 5

]V ert = r I V ert −

[3 2−4 5

]V ert

=

(r I −

[3 2−4 5

])V ert

=

([r 00 r

]−[

3 2−4 5

])V ert

=

[r − 3 −2−(−4) r − 5

]V ert

Plugging this into our model, we find[r − 3 −2

4 r − 5

]V ert =

[00

].



But ert is never 0, so we want r satisfying[r − 3 −2

4 r − 5

]V =

[00

].

For each r , we get two equations in V1 and V2:

(r − 3)V1 − 2V2 = 0, 4V1 + (r − 5)V2 = 0.

Let Ar be this matrix.Any r for which the det Ar 6= 0 tells us these two lines havedifferent slopes and so cross at the origin implying V1 = 0 andV2 = 0. Thus [

xy

]=

[00

]ert =

[00

].

which will not satisfy nonzero initial conditions. So rejectthese r .Any value of r for which det Ar = 0 gives an infinite number ofsolutions which allows us to pick one that matches the initialconditions we have.



The equation

det(r I− A) = det

[r − 3 −2

4 r − 5

]= 0.

is called the characteristic equation of this linear system.

The characteristic equation is a quadratic, so there arethree possibilities:

two distinct roots – this is the only case we handle in this class.the real roots are the same – not done in this class. The thirdCalculus for Biology class does this one.the roots are complex – not done in this class. The thirdCalculus for Biology class does this one.



Examples

Example

Derive the characteristic equation for the system below

x ′(t) = 8 x(t) + 9 y(t)

y ′(t) = 3 x(t) − 2 y(t)

x(0) = 12

y(0) = 4

Solution

The matrix - vector form is[x ′(t)y ′(t)

]=

[8 93 −2

] [x(t)y(t)

].[

x(0)y(0)

]=

[12

4

].



Examples

Solution

The coefficient matrix A is thus

A =

[8 93 −2

].


Plug this into the system.

r V ert −[

8 93 −2

]V ert =

[00

].

Rewrite using the identity matrix I and factor(r I −

[8 93 −2

])V ert =

[00

].



Examples

Solution

Since ert 6= 0 ever, we find r and V satisfy(r I −

[8 93 −2

])V =

[00

].

If r is chosen so that det (rI − A) 6= 0, the only solution tothis system of two linear equations in the two unknowns V1

and V2 is V1 = 0 and V2 = 0. This leads to x(t) = 0 andy(t) = 0 always and this solution does not satisfy the initialconditions. Hence, we must find r which givedet (rI − A) = 0. The characteristic equation is thus

det

[r − 8 −9−3 r + 2

]= (r − 8)(r + 2)− 27

= r2 − 6r − 43 = 0.



Examples

Example

Derive the characteristic equation for the system below

x ′(t) = −10 x(t) − 7 y(t)

y ′(t) = 8 x(t) + 5 y(t)

x(0) = −1

y(0) = −4

Solution

The matrix - vector form is[x ′(t)y ′(t)

]=

[−10 −7

8 5

] [x(t)y(t)

].[

x(0)y(0)

]=

[−1−4

].



Examples

Solution

The coefficient matrix A is thus

A =

[−10 −7

8 5

].


Plug this into the system.

r V ert −[−10 −7

8 5

]V ert =

[00

].

Rewrite using the identity matrix I and factor(r I −

[−10 −7

8 5

])V ert =

[00

].



Examples

Solution

Since ert 6= 0 ever, we find r and V satisfy(r I −

[−10 −7

8 5

])V =

[00

].

If r is chosen so that det (rI − A) 6= 0, the only solution tothis system of two linear equations in the two unknowns V1

and V2 is V1 = 0 and V2 = 0. This leads to x(t) = 0 andy(t) = 0 always and this solution does not satisfy the initialconditions. Hence, we must find r which givedet (rI − A) = 0. The characteristic equation is thus

det

[r + 10 7−8 r − 5

]= (r + 10)(r − 5) + 56

= r2 + 5r + 6 = 0.



Examples

Homework 50

For each of these problems,

Write the matrix - vector form.Derive the characteristic equation but you don’t have to findthe roots.

50.1

x ′ = 2 x + 3 y

y ′ = 8 x − 2 y

x(0) = 3

y(0) = 5.

50.2

x ′ = −4 x + 6 y

y ′ = 9 x + 2 y

x(0) = 4

y(0) = −6.


Finding The General Solution

The roots to the characteristic equation are calledeigenvalues.

Organize the eigenvalues as r1 < r2.

Example: The eigenvalues are −2 and −1. So r1 = −2 andr2 = −1. Since e−2t decays faster than e−t , we say the rootr2 = −1 is the dominant part of the solution.Example: The eigenvalues are −2 and 3. So r1 = −2 andr2 = 3. Since e−2t decays and e3t grows, we say the rootr2 = 3 is the dominant part of the solution.Example: The eigenvalues are 2 and 3. So r1 = 2 and r2 = 3.Since e2t grows slower than e3t , we say the root r2 = 3 is thedominant part of the solution.

For each eigenvalue r we want to find nonzero vectors V sothat (r I − A) V = 0 where to help with our writing we let 0be the two dimensional zero vector.

These nonzero V are called the eigenvectors for eigenvaluer and satisfy AV = rV.



For eigenvalue r1Find V so that (r1 I − A) V = 0There will be an infinite number of V’s that solve this; we pickone and call it eigenvector E1.

For eigenvalue r2Find V so that (r2 I − A) V = 0There will again be an infinite number of V’s that solve this;we pick one and call it eigenvector E2.

The general solution to our model will be[x(t)y(t)

]= AE1 er1t + BE2 er2t .

where A and B are arbitrary. We use the ICs to find A and B.

Best to show all this with some examples.



Examples

Example

For the system below[x ′(t)y ′(t)

]=

[−20 12−13 5

] [x(t)y(t)

][

x(0)y(0)

]=

[−1

2

]

Find the characteristic equation

Find the general solution

Solve the IVP



Examples

Solution

The characteristic equation is

det

(r

[1 00 1

]−[−20 12−13 5

])= 0

Thus

0 = det

([r + 20 −12

13 r − 5

])= (r + 20)(r − 5) + 156

= r2 + 15r + 56

= (r + 8)(r + 7)

Hence, eigenvalues or roots of the characteristic equationare r1 = −8 and r2 = −7.



Examples

First, for eigenvalue r1 = −8, let’s find our eigenvector.

Solution

For eigenvalue r1 = −8, substitute the value into[r + 20 −12

13 r − 5

] [V1

V2

]⇒[

12 −1213 −13

] [V1

V2

]This system of equations should be collinear: i.e. the rowsshould be multiples; i.e. both give rise to the same line. Ourrows are multiples, so we can pick any row to find V2 in termsof V1. Picking the top row, we get 12V1 − 12V2 = 0implying V2 = V1.

Letting V1 = a, we find V1 = a and V2 = a: so[V1

V2

]= a

[11

]



Examples

Solution

Choose E1:

The vector

E1 =

[11

]is our choice for an eigenvector corresponding to eigenvaluer1 = −8.

So one of the solutions is[x1(t)y1(t)

]= E1 e−8t =

[11

]e−8t .



Examples


Solution

For eigenvalue r2 = −7, substitute the value into[r + 20 −12

13 r − 5

] [V1

V2

]⇒[

13 −1213 −12

] [V1

V2

]This system of equations should be collinear: i.e. the rowsshould be multiples; i.e. both give rise to the same line. Ourrows are multiples, so we can pick any row to find V2 in termsof V1. Picking the top row, we get 13V1 − 12V2 = 0implying V2 = (13/12)V1.

Letting V1 = b, we find V1 = b and V2 = (13/12)b: so[V1

V2

]= b

[1

1312

]



Examples

Solution

Choose E2:

The vector

E2 =

[1

1312



]= E2 e−7t =

[1

1312

]e−7t .



Examples

Solution

The general solution:

[x(t)y(t)

]= A E1 e−8t + B E2 e−7t

= A

[11

]e−8t + B

[1

1312

]e−7t .

Find A and B: use the ICs.[x(0)y(0)

]=

[−1

2

]= A

[11

]e0 + B

[1

1312

]e0

= A

[11

]+ B

[1

1312

].



Examples

Solution

So

A + B = −1

A +13

12B = 2

Subtracting the bottom equation from the top equation, weget − 1

12B = −3 or B = 36. Thus, A = −1− B = −37. So[x(t)y(t)

]= −37

[11

]e−8t + 36

[1

1312

]e−7t



Examples

Example

For the system below[x ′(t)y ′(t)

]=

[4 9−1 −6

] [x(t)y(t)

][

x(0)y(0)

]=

[4−2

]

Find the characteristic equation

Find the general solution

Solve the IVP



Examples

Solution

The characteristic equation is

det

(r

[1 00 1

]−[

4 9−1 −6

])= 0

Thus

0 = det

([r − 4 −9

1 r + 6

])= (r − 4)(r + 6) + 9

= r2 + 2r − 15

= (r + 5)(r − 3)

Hence, eigenvalues or roots of the characteristic equationare r1 = −5 and r2 = 3.



Examples


Solution

For eigenvalue r1 = −5, substitute the value into[r − 4 −9

1 r + 6

] [V1

V2

]⇒[−9 −9

1 1

] [V1

V2

]This system of equations should be collinear: i.e. the rowsshould be multiples; i.e. both give rise to the same line. Ourrows are multiples, so we can pick any row to find V2 in termsof V1. Picking the bottom row, we get V1 + V2 = 0 implyingV2 = −V1.

Letting V1 = a, we find V1 = a and V2 = −a: so[V1

V2

]= a

[1−1

]



Examples

Solution

Choose E1:

The vector

E1 =

[1−1



]= E1 e−5t =

[1−1

]e−5t .



Examples

First, for eigenvalue r2 = 3, let’s find our eigenvector.

Solution

For eigenvalue r2 = 3, substitute the value into[r − 4 −9

1 r + 6

] [V1

V2

]⇒[−1 −9

1 9

] [V1

V2

]This system of equations should be collinear: i.e. the rowsshould be multiples; i.e. both give rise to the same line. Ourrows are multiples, so we can pick any row to find V2 in termsof V1. Picking the bottom row, we get V1 + 9V2 = 0implying V2 = −(1/9)V1.

Letting V1 = b, we find V1 = b and V2 = −(1/9)b: so[V1

V2

]= b

[1−1

9

]



Examples

Solution

Choose E2:

The vector

E2 =

[1−1

9

]is our choice for an eigenvector corresponding to eigenvaluer2 = 3.


]= E2 e3t =

[1−1

9

]e3t .



Examples

Solution

The general solution:

[x(t)y(t)

]= A E1 e−5t + B E2 e3t

= A

[1−1

]e−5t + B

[1−1

9

]e3t .

Find A and B: use the ICs.[x(0)y(0)

]=

[4−2

]= A

[1−1

]e0 + B

[1−1

9

]e0

= A

[1−1

]+ B

[1−1

9

].



Examples

Solution

So

A + B = 4

−A− 1

9B = −2

Adding the bottom equation and the top equation, we get89B = 2 or B = 9/4. Thus, A = 4− B = 7/4. So[

x(t)y(t)

]=

7

4

[1−1

]e−5t +

9

4

[1−1

9

]e3t



Examples

Homework 51

For these problems:

write matrix, vector form.

find characteristic equation. No derivation needed this time.

find the two eigenvalues.

find the two associated eigenvectors in glorious detail.

Write general solution.

solve the IVP.

51.1

x ′ = 3 x + y

y ′ = 5 x − y

x(0) = 4

y(0) = −6.

The eigenvalues should be −2 and 4.



Examples

Homework 51 Continued

51.2

x ′ = x + 4 y

y ′ = 5 x + 2 y

x(0) = 4

y(0) = −5.


51.3

x ′ = −3 x + y

y ′ = −4 x + 2 y

x(0) = 1

y(0) = 6.


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