Fundamental Matrix Solution (FMS) - math.utah.edu

Remark and definition: Group the product expression for et A as follows:

et A = P et D P 1

e t A = v1 v2 ... vn

e 1 t

0 ... 0

0 e 2 t

... 0

0 0 ... 0

0 0 ... e n t

v1 v2 ... vn

1

= e 1

tv1 e 2

tv2 ... e n

tvn

v1 v2 ... vn

1

= t 0 1 .In this case we call t a Fundamental Matrix Solution (FMS) for the linear system of differential equations

x t = A x .This is because every solution to the system can be written uniquely as a linear combination of the columns of t , i.e. as t c. We use the same definition in the case that A is not diagonalizable, namely that the columns of t should be a basis for the solution space. And then it will always be true that

et A = t 0 1 .

Exercise 1 Use Theorem 3 to recompute et A for

A = 0 1

1 0.

(This would be even easier if the eigendata was real instead of complex.)

et't III IsIAXII 4,1 72 1 roots a ti

Ea I.li oo span1l Ea span ftAP PD

to et iDp till P

e left eit oit it lit itE L lo e

py XLEleitte it I eit e it

ileit eit eit e.it

eitte it wsttisinttwst isint2wsLeite it cost is int lust is.int 2isint

so entries simplify to f shift

Fri April 55.7 Matrix exponentials as integrating factors for nonhomogeneous systems of linear differential equations.

Announcements:

Warm-up Exercise:

Wolframalpha will compute eat seems reliablefor Aziz

e g et b matrixexponential t 0,13E1.03

We'll focus on today's application

If time finish Exercise 1 Weednah

Generalizing what we did in Chapter 1 for certain scalar first order linear DE's with constant coefficient, to systems, using integrating factors!

Formula for the solution to the IVPx t = A x f t

x 0 = x0 where A is a constant matrix. This is sort of amazing! We'll check every step:

x t = A x f t

x t A x = f t

e t A x t A x = e t Af t

ddt

e t Ax t = e t Af t .

Integrate from 0 to t:

e t Ax t x0 =0

te s Af s ds

Move the x0 over and multiply both sides by et A :

x t = et A x00

te s Af s ds .

This formula can be reorganized a bit, in a way which nicely isolates the homogeneous solution contribution related to the initial data x0 , added to the particular solution which is zero initial data:

x t = e t Ax0 0

te t s Af s ds .

This formula also illustrates a deep fact about linear differential equations and certain partial differential equations as well: The solution to the inhomogeneous DE is determined by the initial conditions, and a so-called convolution integral in which the integrand at time s , with 0 s t, is a linear combination of homogeneous solutions at time t s ( t t s 0) with the forcing function at time s . So, the historyof the forcing before time t, combined with the homogeneous solutions at the specified related but differenttimes, determines the inhomogeneous solution precisely in this particular way.

You can read more about the general Duhamel's Principle at Wikipedia and you might encounter it in moreadvanced applied mathematics or physics courses.

WeddatetB Betts

etBBseederivation

ChptrL

x'It ax f HIPitt atI F e

dat e tax tAY µt eTAI tt

fotddsleSA.is ds eTAI A IiteSAILS e TAI'MetafetA.ity etA Iot

SotesAIGids

I etAfo tetAJoteSAFesidsU

I

Friday application: We love the forced oscillator equation

x t x t = f t .

For today's numerical experiment visualize a swing (pendulum) with the numerical value of gL

= 1. They

used to have swings like that at Liberty Park - super long chains of around 10 m in length. For such a linearized pendulum, i.e. small oscillations from the stable equilibrium, the natural angular frequency is

0 = 1, andxH t = c1 cos t c2sin t

is simple harmonic motion of period 2 6.2 seconds. Happy times.

Now, pretend you're a parent pushing your child on a swing, and you want to teach them about resonance. You did some work before going to the park and know that you don't have to only force with sinusoidal functions any more in order to figure out the response function x t . Besides, who pushes a swing like that anyways? In particular, you checked that if x t solves the second order DE, then

ddt

x t

x t=

x t

x t=

x

x f.

So, x, x T solves the linear systemx1 t

x2 t=

0 1

1 0

x1

x2

0

f.

x t = A x f t

x t = e t Ax0 0

te t s Af s ds .

On Tuesday we computed

et A =cos t sin t

sin t cos t.

linearizedunforcedpendulum

0 t ofO O

I ol lift

wed

Exercise 1 Using the matrix exponential solution formula for the system, and the correspondence back to the second order DE, verify that the solution to the original second order IVP for the swing is

x t = x0 cos t v0 sin t0

tf s sin t s ds

t x f

f it etA Io t g et A f is ds

Hit is c s

t its c inIt xocost t Vosint t J f

lysinH sid s

Any cont fm

> >

We are now ready to play the resonance game. We'll assume x0 = 0, v0 = 0 for simplicity.

Example 1x t x t = sin t

x 0 = 0 x 0 = 0.

with plots : f1 t sin t ; plot1a plot f1 t , t = 0 ..30, color = green : display plot1a, title = `forcing with sinusoidal function at natural period` ;

f1 t sin t

t10 20 301

1forcing with sinusoidal function at natural period

What's your vote? Will we get resonance?

Wo Inatural period To 21T

yes

> > x1 t0

tsin f1 t d :

x1 t ; plot1b plot x1 t , t = 0 ..30, color = black : display plot1a, plot1b , title = `resonance response ?` ;

sin t2

cos t t2

t10 20 30

10

0

10resonance response ?

> >

> >

Example 2) A square wave forcing function with amplitude 1 and period 2 , made with a linear combinations of unit step functions...

with plots :

f2 t 1 2n = 0

10

1 n Heaviside t n Pi :

plot2a plot f2 t , t = 0 ..30, color = green : display plot2a, title = `square wave forcing at natural period` ;

t10 20 301

1square wave forcing at natural period

1) What's your vote? Is this square wave going to induce resonance, i.e. a response with linearly growingamplitude?

sun It II III Z I E t 3IT

i

> > x2 t0

tsin f2 t d :

plot2b plot x2 t , t = 0 ..30, color = black : display plot2a, plot2b , title = `resonance response ?` ;

t10 20 30

100

10

resonance response ?

> >

Example 3) A triangle wave forcing function, same period

f3 t0

tf2 s ds 1.5 : # this antiderivative of square wave should be triangle wave

plot3a plot f3 t , t = 0 ..30, color = green : display plot3a, title = `triangle wave forcing at natural period` ;

10 20 301

triangle wave forcing at natural period

3) Resonance? Yes

> > x3 t0

tsin f3 t d :


t10 20 30

100

10


> >

Example 4) Forcing not at the natural period, e.g. with a square wave having period T = 2 .

f4 t 1 2n = 0

20

1 n Heaviside t n :

plot4a plot f4 t , t = 0 ..20, color = green : display plot4a, title = `periodic forcing, not at the natural period` ;

t5 10 15 201

01

periodic forcing, not at the natural period

4) Resonance? Don't expect resonance since theperiod of forcing 2has no connection to the naturalperiodof2it

> > x4 t0

tsin f4 t d :

plot4b plot x4 t , t = 0 ..20, color = black : display plot4a, plot4b , title = `poor kid ` ;

t5 10 15 20

10.5

00.5

1poor kid

> >

Example 5) Forcing not at the natural period, e.g. with a particular wave having period T = 6 .

f5 tn = 0

10

Heaviside t 6 n Heaviside t 6 n 1 :

plot5a plot f5 t , t = 0 ..150, color = green : display plot5a, title = `sporadic square wave with period 6 ` ;

t50 100 150

01

sporadic square wave with period 6

5) Resonance?

3 naturalperiod

fit to IEEE'IesI 6 I E t ET tO 7it E tE 12 iti

Yes if you thinkabout it We're applyingthe same

push every threecyclesoftheswing

> >

> >

x5 t0

tsin f5 t d :


t50 100 150

1510505

1015


yes

Hey, what happened???? How do we need to modify our thinking if we force a system with somethingwhich is not sinusoidal, in terms of worrying about resonance? In the case that this was modeling a swing(pendulum), how is it getting pushed?

Precise Answer: It turns out that any periodic function with period P is a (possibly infinite)

superposition of a constant function with cosine and sine functions of periods P,P2

,P3

,P4

,... .

Equivalently, these functions in the superposition are

1, cos t , sin t , cos 2 t , sin 2 t , cos 3 t , sin 3 t ,... with ω =2P

. This is the

theory of Fourier series, which we will start studying on Monday next week. If the given periodic forcing function f t has non-zero terms in this superposition for which n = 0 (the natural angular

frequency) (equivalently Pn

=2

0

= T0 ), there will be resonance; otherwise, no resonance.

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Fundamental Matrix Solution (FMS) - math.utah.edu