Higher-order equations as systems

Consider a falling body under the force of near-Earth gravity:

F = ma = mdv

dt

F = −mg

Eliminating F we have

mdv

dt= −mg

dv

dt= −g

System or higher-order?

dv

dt= −g

Let y be the height. Then we can write a higher-order equationusing v = dy/dt:

d2y

dt2= −g

or we can consider the system

dy

dt= v

dv

dt= −g

As a system

d

dt

(

y

v

)

=

(

0 10 0

)(

y

v

)

−

(

0g

)

It’s not a very interesting system as there are only zeroeigenvalues.

◮ One solution is given by integrating the non-homogeneousterm: v = −gt, y = −

1

2gt2.

◮ The homogeneous equation always has v constant, say v0,and y = vt + y0.

◮ Hence the general solution is y = y0 + v0t −1

2gt2

Nothing surprising here, but we see how it fits into this course’sframework.

Starting with a higher-order equation

What if we were just given y ′′ = −g to start with? Then we couldintroduce velocity as y ′ and rewrite the single, second-orderequation as a system:

y ′ = v

v ′ = y ′′ = −g

Now just forget about y ′′ and we have a first-order system:

y ′ = v

v ′ = −g

From one higher-order equation to a first-order systemExample 1 from the textbook:

y ′′ + 3y ′ + 2y = 0

We could introduce v = y ′ and write the equations for

(

y

v

)

or(

v

y

)

. But to follow the textbook’s notation we abandon y

altogether and define

v1 = y

v2 = y ′

Then

v ′

1 = v2

v ′

2 = y ′′ = −2y ′− 2y

v ′

2 = −2v1 − 3v2

As a matrix equation

v ′

1 = v2

v ′

2 = y ′′ = −2y ′− 2y

v ′

2 = −2v1 − 3v2

becomes(

v1

v2

)

′

=

(

0 1−2 −3

)(

v1

v2

)

Solving it as we did before

We solved second-order linear equations before. Let’s apply thatmethod to

y ′′ + 3y ′ + 2y = 0.

We formed the characteristic equation λ2 + 3λ + 2 = 0 and so(λ + 2)(λ + 1) = 0, so λ = −2 or −1 and the solution isc1e

−t + c2e−2t .

Solving it as a system

As a system it has the form

v ′ =

(

0 1−2 −3

)

v

so we need to find the eigenvalues. The equation is

0 =

∣

∣

∣

∣

−λ 1−2 −3 − λ

∣

∣

∣

∣

= λ(3 + λ) + 2

Multiplying out we have λ2 + 3λ + 2 = 0, the same equation as wegot from the higher-order equation, so the characteristic equationof the matrix is the same as the characteristic equation of thehigher-order equation. (That’s why it’s OK to use the same words“characteristic equation” in both situations.)

Find the eigenvectors

I choose to expand on the second row. So, to find theeigenvector’s first element I ignore the first column (and secondrow), which leaves the upper right element 1. To find the secondelement I ignore the second column (and second row), leaving theupper left element −λ. But I have to change the sign, because thesecond column is an even-numbered column. So the eigenvector

for λ is

(

1λ

)

. Putting λ = −1 and then −2 we have

v−1 =

(

1−1

)

v−2 =

(

1−2

)

Compute the change-of-basis matrix P

Put the eigenvalues in the columns:

P =

(

1 1−1 −2

)

The first column is v−1 and the second column is v−2.

Find the solution

So the solution is

v = P

(

c1e−t

c2e−2t

)

=

(

1 1−1 −2

)(

c1e−t

c2e−2t

)

=

(

c1e−t + c2e

−2t

−c1e−t

− 2c2e−2t

)

Since y = v1 that gives us

y = c1e−t + c2e

−2t

Sure enough, just what we could have gotten a month ago withless trouble!

So what?!

◮ It may seem foolish to work harder to get the same answer wecould get more easily before.

◮ But we can now solve problems we could not solve before.

◮ For example, given a system of three second-order equations,now we can convert it two a system of six first-order equation,which is one first-order matrix equation.

◮ That would come up solving the problem of an object movingin three dimensions under some forces (such as gravity and/orrockets).

◮ Yes, this is rocket science!

Non-constant coefficients

Moreover we did not have a method for solving even second-orderequations with non-constant coefficients.

◮ Not that we have one for systems either

◮ but numerical methods always are written for first-ordersystems, so this reduction method is very important, becauseyou have to reduce your problem to a first-order system beforeusing a computer on it.

Order higher than second

y ′′′ + 4y ′′− y ′

− 4y = 0

Now we have three variables v1 = y , v2 = y ′, v3 = y ′′ = v ′

2and we

get

v1

v2

v3

′

=

0 1 00 0 14 1 −4

v = 0

Generally you will get 1s on the “super diagonal” (one above themain diagonal) and then the bottom row will contain thecoefficients of the original equation (if the first coefficient, of thehighest derivative, is 1).

The characteristic equation

y ′′′ + 4y ′′− y ′

− 4y = 0

The characteristic equation of this equation is

λ3 + 4λ2− λ − 4 = 0

The characteristic equation of the matrix is

∣

∣

∣

∣

∣

∣

−λ 1 00 −λ 14 1 −4 − λ

∣

∣

∣

∣

∣

∣

= −λ3− 4λ2 + λ + 4

so the general solution is going to involve the same exponentialseλt , whether I solve it the old way or as a system.

Let’s solve this example

λ3 + 4λ2− λ − 4 = 0

First we need the eigenvalues. It’s a cubic but it’s a textbookexample, so likely at least one of the eigenvalues divides 4. Well, 4is too big, and 2 is too big, but λ = 1 works, so we can divide byλ − 1 to get a quadratic equation, λ2 + 5λ + 4 = 0. That factorsinto (λ + 4)(λ + 1), so the eigenvalues are 1, −1, and −4.

Old method solution

So by the old methods, the solution is

y = c1et + c2e

−t + c3e−4t

But to finish by the new method, we need to find the matrix P

whose columns are the eigenvectors and multiply

v =

c1et

c2e−t

c3e−4t

by that matrix P . However, the solution for y is going to just bethe first row of P times v , so it will also be just a linearcombination of those same exponentials, unless possibly some ofthe entries on that row of P are zero; that is, if some of theeigenvectors have zero first component.

Those eigenvectorsWhat are the eigenvectors of the matrix? Look at

−λ 1 00 −λ 14 1 −4 − λ

= −λ3− 4λ2 + λ + 4

I will show you that the eigenvectors never have zero firstcomponent. To find the eigenvectors I choose to expand on thelast row (because there are a lot of zeroes elsewhere). The firstelement of the eigenvector for λ is found by ignoring the last rowand the first column. That leaves a two-by-two determinant givingthe first component:

∣

∣

∣

∣

1 0−λ 1

∣

∣

∣

∣

This evaluates to 1, so it’s not zero.Hence the general solution is just a linear combination of theexponential functions determined by the eigenvalues–just what wegot by the old method.

Order higher than three

In general the same thing will happen: The first component of theeigenvector will be given by an n − 1 by n − 1 determinant that hasthe form (shown for n = 4).

∣

∣

∣

∣

∣

∣

1 0 0−λ 1 00 −λ 1

∣

∣

∣

∣

∣

∣

Since it’s a triangular matrix, its determinant is the product of thediagonal elements, which is 1.

So what?

◮ So the methods we learned for systems give the same resultson higher-order equations as we could already get, and

◮ They also work on systems of higher-order equations, and

◮ They also enable us to reduce higher-order to equations tofirst-order systems, even if the coefficients are not constant

Example 3 from textbook

y ′′− y ′

− 2y = sin x

v1 = y

v2 = y ′

v ′

1 = v2

v ′

2 = y ′′ = y ′ + 2y + sin x = 2v1 + v2 + sin x

v ′ =

(

0 12 1

)

v +

(

0sin x

)

Eigenvalues

Find eigenvalues and eigenvectors of

(

0 12 1

)

We have

0 =

∣

∣

∣

∣

−λ 12 1 − λ

∣

∣

∣

∣

0 = λ2− λ − 2 = (λ − 2)(λ + 1)

so the eigenvalues are λ = −1 and 2.

Find eigenvectors

Consider(

−λ 12 1 − λ

)

We expand on the bottom row. The first component of theeigenvector is found by ignoring the bottom row and the firstcolumn. So it’s 1. The second component is found by ignoring thebottom row and second column, and changing the sign, so it’s λ.Hence the eigenvector is (1, λ). Putting λ = −1 and then 2 we getthe matrix of eigenvectors,

P =

(

1 1−1 2

)

Find YH

Since

P =

(

1 1−1 2

)

where the columns correspond to λ = −1 and then 2, the solutionof the homogeneous equation is

YH = P

(

c1e−x

c2e2x

)

=

(

1 1−1 2

)(

e−x

e2x

)

=

(

c1e−x + c2e

2x

−c1e−x + 2c2e

2x

)

Make a matrix of linearly independent solutions

v =

(

c1e−x + c2e

2x

−c1e−x + 2c2e

2x

)

Taking (c1, c2) to be first (1, 0) and then (0, 1) we get

M =

(

e−x e2x

−e−x 2e2x

)

We will also need

M−1 =1

3

(

2ex−ex

e−2x e−2x

)

Find one solution of the nonhomogeneous equation

We have

G =

(

0sin x

)

and

M =

(

e−x e2x

−ex 2e2x

)

and

M−1 =1

3

(

2ex−ex

e−2x e−2x

)

so

M−1G =1

3

(

2ex−ex

e−2x e−2x

)(

0sin x

)

=1

3

(

−ex sin x

e−2x sin x

)

Now compute∫

M−1G dx

∫

M−1G dx =1

3

( ∫

−ex sin x dx

e−2x sin x dx

)

These integrals are done by integrating by parts twice, and thensolving for the original integral. You get the following results:

∫

M−1G dx =

(

1

6ex(cos x − sin x)

−

1

15e−2x(cos x + 2 sin x)

)

Find the solution

YP = M

∫

M−1G dx

=

(

e−x e2x

−ex 2e2x

)(

1

6ex(cos x − sin x)

−

1

15e−2x(cos x + 2 sin x)

)

=

(

1

10cos x −

3

10sin x

−

3

10cos x −

1

10sin x

)

So the general solution YH + YP is

Y =

(

c1e−x + c2e

2x + 1

10cos x −

3

10sin x

−c1ex + 2c2e

2x− e−x

−

3

10cos x −

1

10sin x

)

Since v1 = y the solution to the second-order equation we startedwith is

y = c1e−x + c2e

2x +1

10cos x −

3

10sin x